Lanjutan Contoh Soal 2 Persamaan dan Pertidaksamaan Nilai Mutlak dari Bentuk Linear Satu Variabel (Kelas X Wajib)

\begin{array}{ll}\\ \fbox{21}.&\textbf{(USM UGM Mat IPA)}\textrm{Semua nilai \textit{x} yang memenuhi}\: \: x\left | x-2 \right |<x-2\: \: \textrm{adalah... .}\\ &\textrm{a}.\quad x<-1\: \: \textrm{atau}\: \: 1<x<2\\ &\textrm{b}.\quad x<-2\\ &\textrm{c}.\quad -2<x<-1\\ &\textrm{d}.\quad x<-1\\ &\textrm{e}.\quad -2<x<1\\\\ &\textrm{Jawab}:\quad \textbf{d}\\ &\begin{array}{|c|c|}\hline \multicolumn{2}{|c|}{\begin{aligned}&\\ &x\left | x-2 \right |<x-2\\ & \end{aligned}}\\\hline x<2&x\geq 2\\\hline \begin{aligned}x\left | x-2 \right |&<x-2\\ x(2-x)&<x-2\\ 2x-x^{2}&<x-2\\ -x^{2}+x+2&<0\\ x^{2}-x-2&>0\\ (x-2)(x+1)&>0\end{aligned}&\begin{aligned}x\left | x-2 \right |&<x-2\\ x(x-2)&<x-2\\ x^{2}-2x&<x-2\\ x^{2}-3x+2&<0\\ (x-1)(x-2)&<0\\ & \end{aligned}\\ \begin{array}{ccc|cccc|cccccc}\\ &&\multicolumn{2}{c}{.}&&&\multicolumn{2}{c}{.}\\\cline{1-3}\cline{8-9} +&+&&-&-&-&-&+&+&\quad \textbf{X}\\\hline &&\multicolumn{2}{l}{-1}&&&\multicolumn{2}{c}{2}&\\ \end{array}&\begin{array}{ccc|cccc|cccccc}\\ &&\multicolumn{2}{c}{.}&&&\multicolumn{2}{c}{.}\\\cline{4-7} +&+&&-&-&-&-&+&+&\quad \textbf{X}\\\hline &&\multicolumn{2}{c}{1}&&&\multicolumn{2}{c}{2}&\\ \end{array} \\\hline \textbf{ada yang memenuhi}&\textbf{tidak ada yang memenuhi}\\\hline \multicolumn{2}{|c|}{\begin{aligned}\textrm{yang memenuhi}:&\\ &x<-1\\ & \end{aligned}}\\\hline \end{array} \end{array}.

Sumber Referensi

  1. Budhi, W. S. 2014. Bupena Matematika SMA/MA Kelas X Kelompok Wajib. Jakarta: Erlangga.
  2. Susianto, B. 2011. Olimpiade Matematika dengan Proses Berpikir Aljabar dan Bilangan. Jakarta: Grasindo.
  3. Yuana, R. A., Indriyastuti. 2017. Perspektif Matematika 1 untuk Kelas X SMA dan MA Kelompok Mata Pelajaran Wajib. Solo: PT. Tiga Serangkai Pustaka Mandiri.
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Lanjutan Contoh Soal Persamaan dan Pertidaksamaan Nilai Mutlak dari Bentuk Linear Satu Variabel (Kelas X Wajib)

\begin{array}{ll}\\ \fbox{11}.&\textrm{Nilai}\: \: k\: \: \textrm{yang memenuhi}\: \: \left | 4k \right |=16\: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \textrm{a}.&2\\ \textrm{b}.&4\\ \textrm{c}.&\pm 2\\ \textrm{d}.&\pm 4\\ \textrm{e}.&\pm 8 \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{d}\\ &\begin{aligned}\left | 4k \right |&=16\\ (4k)&=\pm 16\\ k&=\pm \displaystyle \frac{16}{4}\\ &=\pm 4 \end{aligned}\end{array}.

\begin{array}{ll}\\ \fbox{12}.&\textrm{Nilai}\: \: p\: \: \textrm{yang memenuhi}\: \: 10-4\left | 4-5p \right |=26\: \: \textrm{adalah... .}\\ &\begin{array}{lllllll}\\ \textrm{a}.&2\: \: \textrm{atau}\: \: 1\displaystyle \frac{2}{3}&&&\textrm{d}.&1\: \: \textrm{atau}\: \: -\displaystyle \frac{3}{5}\\\\ \textrm{b}.&2\: \: \textrm{atau}\: \: 2\displaystyle \frac{3}{5}\quad&\textrm{c}.&-2\displaystyle \frac{3}{4}\: \: \textrm{atau}\: \: 1\quad&\textrm{e}.&-1\: \: \textrm{atau}\: \: 2\displaystyle \frac{3}{5}\\ \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{e}\\ &\begin{aligned}10-4\left | 4-5p \right |&=-26\\ -4\left | 4-5p \right |&=-36\\ \left | 4-5p \right |&=9\\ (4-5p)&=\pm 9\\ -5p&=-4\pm 9\\ p&=\displaystyle \frac{-4\pm 9}{-5}\\ p&=\begin{cases} \displaystyle \frac{-4+9}{-5} & =-1 \\ \textrm{atau} & \\ \displaystyle \frac{-4-9}{-5} & = \displaystyle \frac{13}{5}=2\frac{3}{5} \end{cases} \end{aligned} \end{array}.

\begin{array}{ll}\\ \fbox{13}.&\textrm{Perhatikanlah ilustrasi grafik di bawah ini}\\ &\begin{aligned}\end{aligned}\end{array}.

\begin{array}{ll}\\ .\: \: \: \: \: \: &\textrm{Persamaan yang memenuhi rumus tersebut adalah... .}\\ &\begin{array}{llll}\\ \textrm{a}.&y=\left | -x-2 \right |\\ \textrm{b}.&y=-2x-4\\ \textrm{c}.&y=-\left | 2x-4 \right |\\ \textrm{d}.&y=\left | -2x-4 \right |\\ \textrm{e}.&y=\left | -2x+4 \right | \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{e}\\ &\begin{aligned}&\textrm{Dengan cara substitusi langsung kita akan mendapatkan}\\ &\bullet \quad \textrm{untuk}\: \: x=4\: \: \textrm{menyebabkan nilai}\: \: y=4\: \: \textrm{dan}\\ &\qquad \textrm{sampai langkah di sini hanya ada 1 persamaan yang memenuhi yaitu}:\: \: y=\left | -2x+4 \right | \end{aligned}\end{array}.

\begin{array}{lll}\\ \fbox{14}.&\textrm{Gambarlah garfik untuk persamaan}\: \: \left | x \right |+\left | y \right |=4\\\\\\ &\textrm{Jawab}:\\\\ &\begin{array}{|cc|cc|}\hline \multicolumn{4}{|c|}{\begin{aligned}&\\ &\left | x \right |+\left | y \right |=4\\ & \end{aligned}}\\\hline x> 0\: ,\: y> 0&&&x> 0\: ,\: y<0\\\hline x+y=4&&&x+(-y)=4\\ &&&\\\hline &&&\\ x<0\: ,\: y> 0&&&x<0\: ,\: y<0\\\hline (-x)+y=4&&&(-x)+(-y)=4\\\hline \end{array} \end{array}.

Berikut ini untuk ilustrasi grafiknya

\begin{array}{lll}\\ \fbox{15}.&\textrm{Tentukanlah nilai}\: \: x\: \: \textrm{dan}\: \: y\: \: \textrm{yang memenuhi}\: \: x+\left | x \right |+y=5\: \: \textrm{dan}\: \: x+\left | y \right |-y=10\\\\ &\textrm{Jawab}:\\\\ &\begin{array}{|cc|cc|}\hline \multicolumn{4}{|c|}{\begin{aligned}&\\ &\left | x \right |+x+y=5\\ &\: \: \qquad \textrm{dan}\\ &x+\left | y \right |-y=10\\ & \end{aligned}}\\\hline x> 0\: ,\: y> 0\quad \textbf{(kuadran I)}&&&x> 0\: ,\: y<0\quad \textbf{(kuadran IV)}\\\hline \begin{cases} \left | x \right |+x+y=5 & \Leftrightarrow (x)+x+y=5 \\ &\Leftrightarrow 2x+y=5\: (\textrm{ada})\\ x+\left | y \right |-y=10 & \Leftrightarrow x+(y)-y=10\\ &\Leftrightarrow x=10\: (\textrm{ada}) \end{cases}&&&\begin{cases} \left | x \right |+x+y=5 & \Leftrightarrow (x)+x+y=5\\ &\Leftrightarrow 2x+y=5\: (\textrm{ada})\\ x+\left | y \right |-y=10 & \Leftrightarrow x+(-y)-y=10\\ &\Leftrightarrow x-2y=10\: (\textrm{ada}) \end{cases}\\ &&&\\\hline &&&\\ x<0\: ,\: y> 0\quad \textbf{(kuadran II)}&&&x<0\: ,\: y<0\quad \textbf{(kuadran III)}\\\hline \begin{cases} \left | x+x+y=5 \right | & \Leftrightarrow (-x)+x+y=5\\ &\Leftrightarrow y=5\: (\textrm{ada})\\ x+\left | y \right |-y=10 & \Leftrightarrow x+(y)-y=10\\ &\Leftrightarrow x=10\: \textbf{(tidak memenuhi)} \end{cases}&&&\begin{cases} \left | x \right |+x+y=5 & \Leftrightarrow (-x)+x+y=5\\ &\Leftrightarrow y=5\: \textbf{(tidak memenuhi)}\\ x+\left | y \right |-y=10 & \Leftrightarrow x+(-y)-y=10\\ &\Leftrightarrow x-2y=10\: (\textrm{ada}) \end{cases}\\\hline \end{array} \end{array}..

Berikut adalah ilustrasi grafiknya 

\begin{array}{ll}\\ \fbox{16}.&\textrm{Gambarlah grafik fungsi mutlak dari}\\ &\textrm{a}.\quad y=\left | x-2 \right |\\ &\textrm{b}.\quad y=-\left | x-2 \right |\\ &\textrm{c}.\quad y=2+\left | x-2 \right |\\ &\textrm{d}.\quad y=2-\left | x-2 \right |\\ &\textrm{e}.\quad y=\left | 2+\left | x-2 \right | \right |\\ &\textrm{f}.\quad y=\left | 2-\left | x-2 \right | \right | \end{array}.

Jawab:

Untuk proses jawaban diserahkan kepada pembaca

berikut beberapa hasil ilustrasi grafiknya.

Untuk jawaban a)

Untuk Jawaban b)

Untuk jawaban c)

Untuk jawaban d)

Untuk jawaban e) silahkan gambar sendiri dan simpulkan sendiri

Untuk Jawaban f)

\begin{array}{ll}\\ \fbox{17}.&\textrm{Seluruh bilangan bilangan real}\: \: x\: \: \textrm{yang jaraknya terhadap 3 kurang dari 1 adalah... .}\\ &\begin{array}{lllllll}\\ \textrm{a}.&3<x<4&&&\textrm{d}.&3<x<5\\ \textrm{b}.&2<x<3 \quad&\textrm{c}.&2<x<4 \quad&\textrm{e}.&1<x<3\\ \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{c}\\ &\begin{aligned}\textrm{Seluruh }&\textrm{bilangan bilangan real}\: \: x\: \: \textrm{yang jaraknya terhadap 3 kurang dari 1, maksudnya adalah:}\\ &\left | x-3 \right |<1\\ -1&<x-3<1\\ -1+\textbf{(3)}&<x-3+\textbf{(3)}<1+\textbf{(3)}\\ 2&<x<4 \end{aligned} \end{array}.

\begin{array}{ll}\\ \fbox{18}.&\textrm{Pernyataan berikut yang tepat adalah... .}\\ &\begin{array}{llll}\\ \textrm{a}.&\textrm{Semua nilai}\: \: m\: \: \textrm{yang memenuhi}\: \: -1<m<2\: \: \textrm{akan memenuhi juga}\: \: \left | m \right |<2\\ \textrm{b}.&\textrm{Semua nilai}\: \: m\: \: \textrm{yang memenuhi}\: \: -1<m<2\: \: \textrm{akan memenuhi juga}\: \: \left | m \right |<1\\ \textrm{c}.&\textrm{Semua nilai}\: \: m\: \: \textrm{yang memenuhi}\: \: -3<m<-2\: \: \textrm{akan memenuhi juga}\: \: \left | m \right |<2\\ \textrm{d}.&\textrm{Semua nilai}\: \: m\: \: \textrm{yang memenuhi}\: \: 2<m<3\: \: \textrm{akan memenuhi juga}\: \: \left | m \right |<2\\ \textrm{e}.&\textrm{pilihan jawaban baik a, b, c, maupun d tidak ada yang benar} \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{a}\\ &\begin{aligned}\textrm{Perhat}&\textrm{ikanlah opsi}\: \: \textbf{a}\: ,\\ \left | m \right |&<2\\ -2<m&<2\\ \textrm{sehing}&\textrm{ga untuk nilai}\: \: m\in \mathbb{R}\: \: \textrm{pada rentang}\\ -1<m&<2\: \: \: \: \textrm{akan memenuhi semua}\\\\ \textbf{silahk}&\textbf{an cek sendiri untuk opsi jawaban yang lain} \end{aligned} \end{array}.

\begin{array}{ll}\\ \fbox{19}.&\textrm{Himpunan penyelesaian dari pertidaksamaan}\: \: \left | x+3 \right |<2\left | x-4 \right |\: \: \textrm{adalah... .}\\ &\begin{array}{lllllll}\\ \textrm{a}.&\left \{ x|x<-\displaystyle \frac{5}{3} \right \}&&&\textrm{d}.&\left \{ x|x<\displaystyle \frac{5}{3} \right \}\cup \left \{ x|x>11 \right \}\\\\ \textrm{b}.&\left \{ x|\: \displaystyle \frac{5}{3}<x<-11 \right \} \quad&\textrm{c}.&\left \{ x|x\geq -11 \right \} \quad&\textrm{e}.&\left \{ x|x>-\displaystyle \frac{5}{3} \right \}\cup \left \{ x|x<-11 \right \} \\ \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{d}\\ &\begin{aligned}\left | x+3 \right |&<2\left | x-4 \right |\\ \left ( x+3 \right )^{2}&<2^{2}\left ( x-4 \right )^{2}\quad \textrm{dikuadratkan masing-masing ruas}\\ x^{2}+6x+9&<4\left ( x^{2}-8x+16 \right )\\ x^{2}-4x^{2}+6x+32x+9-64&<0\\ -3x^{2}+38x-55&<0\\ 3x^{2}-38x+55&>0\\ \left ( 3x-5 \right )\left (x -11 \right )&>0\\\\ \textrm{Berikut untuk}&\: \textrm{garis bilangannya}\\ &\begin{array}{ccc|cccc|cccccc}\\ &&\multicolumn{2}{c}{.}&&&\multicolumn{2}{c}{.}\\\cline{1-3}\cline{8-9} +&+&&&&&&+&+&\quad \textbf{X}\\\hline &&\multicolumn{2}{r}{\begin{matrix} \displaystyle \frac{3}{5}\\ \end{matrix}}&&&\multicolumn{2}{l}{11}&\\ \end{array} \end{aligned} \end{array}.

\begin{array}{ll}\\ \fbox{20}.&\textrm{Himpunan penyelesaian dari pertidaksamaan}\: \: \left | x^{2}+5x \right |\leq 6\: \: \textrm{adalah... .}\\ &\textrm{a}.\quad \left \{ x|-6\leq x\leq 1 \right \}\\ &\textrm{b}.\quad \left \{ x|-3\leq x\leq -2 \right \}\\ &\textrm{c}.\quad \left \{ x|-6\leq x\leq -3 \: \: \textrm{atau}\: \: -2\leq x\leq 1\right \}\\ &\textrm{d}.\quad \left \{ x|-6\leq x\leq -5 \: \: \textrm{atau}\: \: -2\leq x\leq 0\right \}\\ &\textrm{e}.\quad \left \{ x|-5\leq x\leq -3 \: \: \textrm{atau}\: \: -2\leq x\leq 0\right \}\\\\ &\textrm{Jawab}:\quad \textbf{c}\\ &\begin{array}{|c|c|}\hline \multicolumn{2}{|c|}{\begin{aligned}&\\ \left | x^{2}+5x \right |&\leq 6\\ -6\leq x^{2}+5x&\leq 6\\ & \end{aligned}}\\\hline -6\leq x^{2}+5x&x^{2}+5x\leq 6\\\hline \begin{aligned}x^{2}+5x+6&\geq 0\\ (x+3)(x+2)&\geq 0 \end{aligned}&\begin{aligned}x^{2}+5x-6&\leq 0\\ (x+6)(x-1)&\leq 0 \end{aligned}\\\hline &\\ \begin{array}{ccc|cccc|cccccc}\\ &&\multicolumn{2}{c}{.}&&&\multicolumn{2}{c}{.}\\\cline{1-3}\cline{8-9} +&+&&-&-&-&-&+&+&\quad \textbf{X}\\\hline &&\multicolumn{2}{l}{-3}&&&\multicolumn{2}{r}{-2}&\\ \end{array}&\begin{array}{ccc|cccc|cccccc}\\ &&\multicolumn{2}{c}{.}&&&\multicolumn{2}{c}{.}\\\cline{4-7} +&+&&-&-&-&-&+&+&\quad \textbf{X}\\\hline &&\multicolumn{2}{l}{-6}&&&\multicolumn{2}{c}{1}&\\ \end{array} \\ &\\\hline \multicolumn{2}{|c|}{\begin{aligned}\textrm{Kesimpulan}:&\\ &\begin{array}{ccc|cccc|ccc|ccc|cccccccc}\\ &&\multicolumn{2}{l}{.}&&&\multicolumn{2}{c}{.}&&\multicolumn{2}{l}{.}&&\multicolumn{2}{r}{.}\\\cline{4-7}\cline{11-13} &&&&&&&&&&&&&&\quad \textbf{X}\\\hline &&\multicolumn{2}{l}{-6}&&&\multicolumn{2}{r}{-3}&&\multicolumn{2}{l}{-2}&&\multicolumn{2}{l}{1}&\\ \end{array} \\ & \end{aligned}}\\\hline \end{array} \end{array}.

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Contoh Soal Persamaan dan Pertidaksamaan Nilai Mutlak dari Bentuk Linear Satu Variabel (Kelas X Wajib)

\begin{array}{ll}\\ \fbox{1}.&\textrm{Nilai untuk}\: \: -\left | -2019 \right |=...\: .\\ &\begin{array}{llllll}\\ \textrm{a}. &-2019\quad &&&\textrm{d}.&-2019^{-1}\\ \textrm{b}.&2019&\textrm{c}.&2019^{-1}\quad &\textrm{e}.&2019^{2} \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{a}\\ &\begin{aligned}-\left | -2019 \right |&=-\left ( 2019 \right )=-2019 \end{aligned} \end{array}.

\begin{array}{ll}\\ \fbox{2}.&\textrm{Nilai untuk}\: \: \left | -4 \right |-\left | -6^{2}\times 2 \right |=...\: .\\ &\begin{array}{llllll}\\ \textrm{a}. &-68\quad &&&\textrm{d}.&68\\ \textrm{b}.&-40&\textrm{c}.&40\quad &\textrm{e}.&76 \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{a}\\ &\begin{aligned}\left | -4 \right |-\left | -6^{2}\times 2 \right |&=\left ( 4 \right )-\left | -36\times 2 \right |\\ &=4-\left | -72 \right |\\ &=4-\left ( 72 \right )\\ &=-68 \end{aligned} \end{array}.

\begin{array}{ll}\\ \fbox{3}.&\textrm{Jika diberikan}\: \: m\: \: \textrm{adalah bilangan positif dan}\: \: n\: \: \textrm{adalah sebuah bilangan negatif},\\ &\textrm{maka operasi berikut yang mengahsilkan bilangan negatif adalah... .}\\ &\begin{array}{llllll}\\ \textrm{a}. &\left | m\times n \right |\quad &&&\textrm{d}.&m+\left | n \right |\\ \textrm{b}.&m\times \left | n \right |&\textrm{c}.&\left | m \right |\times n\quad &\textrm{e}.&\left | m \right |-n \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{c}\\ &\begin{aligned}&\textrm{Misalkan}\begin{cases} m & = p\\ n &= -q \end{cases}\\ & \begin{array}{|c|c|c|}\hline \textrm{No}&\textrm{Pernyataan}&\textrm{Keterangan}\\\hline \textrm{a}&\left | m\times n \right |=\left | p\times -q \right |=p\times q&\textrm{Positif}\\\hline \textrm{b}&m\times \left | n \right |=p\times \left | -q \right |=p\times q&\textrm{positif}\\\hline \textrm{c}&\left | m \right |\times n=\left | p \right |\times -q=-p\times q&\textbf{negatif}\\\hline \textrm{d}&m+\left | n \right |=p+\left | -q \right |=p+q&\textrm{positif}\\\hline \textrm{e}&\left | m \right |-n=\left | p \right |-(-q)=p+q&\textrm{positif}\\\hline \end{array} \end{aligned} \end{array}.

\begin{array}{ll}\\ \fbox{4}.&\textrm{Bentuk dari}\: \: \displaystyle \frac{1}{\left | b \right |-1}\: \: \textrm{memiliki harga untuk}....\\ &\begin{array}{ll}\\ \textrm{a}. &\textrm{semua bilangan}\: \: b\\ \textrm{b}.&\textrm{semua bilangan}\: \: b\: \: \textrm{kecuali 0}\\ \textrm{c}.&\textrm{semua bilangan positif}\: \: b\\ \textrm{d}.&\textrm{semua bilangan}\: \: b\: \: \textrm{kecuali 1}\\ \textrm{e}.&\textrm{semua bilangan}\: \: b\: \: \textrm{kecuali 1 dan -1} \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{e}\\ &\begin{aligned}\textrm{Bentuk pecahan}\: &\: \displaystyle \frac{1}{\left | b \right |-1}\: \: \textrm{selalu mensyaratkan penyebut}\: \: \neq 0\\ \textrm{Sehingga}\qquad\quad\: \, &\\ \left | b \right |-1&\neq 0\\ \left | b \right |&\neq 1\\ \sqrt{b^{2}}&\neq 1,\quad (\textrm{dikuadratkan masing-masing ruas})\\ b^{2}&\neq 1\\ b^{2}-1&\neq 0\\ (b+1)(b-1)&\neq 0\\ b\neq -1\quad \textrm{atau}\quad b&\neq 1 \end{aligned} \end{array}.

\begin{array}{ll}\\ \fbox{5}.&\textrm{Persamaan nilai mutlak berikut yang bernilai benar adalah... .}\\ &\begin{array}{lll}\\ \textrm{a}. &\left | 2^{3}-3^{2} \right |=3^{2}-2^{3}\\ \textrm{b}.&\left | 3^{4}-4^{3} \right |=4^{3}-3^{4}\\ \textrm{c}.&\left | 4^{5}-5^{4} \right |=5^{4}-4^{5}\\ \textrm{d}.&\left | 5^{6}-6^{5} \right |=6^{5}-5^{6}\\ \textrm{e}.&\textrm{pilihan jawaban a, b, c, dan d tidak ada yang benar} \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{a}\\ &\begin{array}{|c|l|c|}\hline \textrm{No}&\qquad\qquad\qquad\qquad\qquad\qquad \textrm{Pernyataan}&\textrm{Keterangan}\\\hline \textrm{a}&\left | 2^{3}-3^{2} \right |=\left | 8-9 \right |=9-8=3^{2}-2^{3}&\textbf{benar}\\\hline \textrm{b}&\left | 3^{4}-4^{3} \right |=\left | 81-64 \right |=81-64=3^{4}-4^{3}\neq 4^{3}-3^{4}&\textrm{salah}\\\hline \textrm{c}&\left | 4^{5}-5^{4} \right |=\left | 1024-625 \right |=1024-625=4^{5}-5^{4}\neq 5^{4}-4^{5}&\textrm{salah}\\\hline \textrm{d}&\left | 5^{6}-6^{5} \right |=\left | 15625-7776 \right |=15625-7776=5^{6}-6^{5}\neq 6^{5}-5^{6}&\textrm{salah}\\\hline \end{array} \end{array}.

\begin{array}{ll}\\ \fbox{6}.&\textrm{Pernyataan berikut yang benar adalah... .}\\ &\begin{array}{llll}\\ \textrm{a}.&\left | m \right |+\left | -m \right |=0\\ \textrm{b}.&\displaystyle \frac{\left | -6 \right |+2\left | 3 \right |}{6}=0\\ \textrm{c}.&\left |7^{2}\times 2 \right |-\left | -3^{2} \right |=107\\ \textrm{d}.&\left | x \right |=3,\: \: \textrm{hanya dapat dipenuhi oleh}\: \: -3\: \: \textrm{dan}\: \: 3\\ \textrm{e}.&\textrm{tanda nilai mutlak berlaku hanya untuk bilangan positif} \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{d}\\ &\begin{array}{|c|l|c|}\hline \textrm{No}&\qquad\qquad\qquad\qquad\qquad \textrm{Pernyataan}&\textrm{Keterangan}\\\hline \textrm{a}&\left | m \right |+\left | -m \right |=m+m=2m\neq 0&\textrm{salah}\\\hline \textrm{b}&\displaystyle \frac{\left | -6 \right |+2\left | 3 \right |}{6}=\displaystyle \frac{6+6}{6}=2\neq 0&\textrm{salah}\\\hline \textrm{c}&\left |7^{2}\times 2 \right |-\left | -3^{2} \right |=98-9=89\neq 107&\textrm{salah}\\\hline \textrm{d}&\left | x \right |=3,\: \: \textrm{hanya dapat dipenuhi oleh}\: \: -3\: \: \textrm{dan}\: \: 3&\textbf{benar}\\\hline \textrm{e}&\textrm{tanda nilai mutlak berlaku hanya untuk bilangan positif}&\textrm{salah}\\\hline \end{array} \end{array}.

\begin{array}{ll}\\ \fbox{7}.&\textrm{Nilai dari}\: \: \left | 2\pi -5 \right |-\displaystyle \frac{\pi }{2}=...\: .\\ &\begin{array}{llll}\\ \textrm{a}.&\displaystyle \frac{3\pi -10}{5}\\ \textrm{b}.&\displaystyle \frac{3\pi +5}{2}\\ \textrm{c}.&\displaystyle \frac{-3\pi +5}{2}\\ \textrm{d}.&\displaystyle \frac{-3\pi +10}{2}\\ \textrm{e}.&\displaystyle \frac{3\pi -10}{2} \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{e}\\ &\begin{aligned}\left | 2\pi -5 \right |-\displaystyle \frac{\pi }{2}&=\left (2\pi -5 \right )-\displaystyle \frac{\pi }{2},\qquad \textrm{sebagai catatan bahwa}\: \: \: \pi =3,14...\\ &=\displaystyle \frac{2(2\pi )-2(5)-\pi }{2}\\ &=\displaystyle \frac{3\pi -10}{2} \end{aligned} \end{array}.

\begin{array}{ll}\\ \fbox{8}.&\textrm{Bentuk singkat dari}\: \: m-6n\: \: \textrm{atau}\: \: 6n-m\: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \textrm{a}.&\left | 6n+m \right |\\ \textrm{b}.&\left | n-6m \right |\\ \textrm{c}.&\left | m-6n \right |\\ \textrm{d}.&\left | m-n \right |\\ \textrm{e}.&\left | -6n-m \right | \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{c}\\ &\begin{aligned}\left | m-6n \right |=\begin{cases} m-6n & ,\textrm{untuk}\: \: m-6n\geq 0 \\ -(m-6n)=6n-m & ,\textrm{untuk}\: \: m-6n< 0 \end{cases} \end{aligned}\end{array}.

\begin{array}{ll}\\ \fbox{9}.&\textrm{Nilai untuk}\: \: x\: \: \textrm{yang memenuhi persamaan}\: \: \left | 2x-5 \right |=11\: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \textrm{a}.&\textrm{hanya}\: \: 3\\ \textrm{b}.&\textrm{hanya}\: \: 8\\ \textrm{c}.&-3\: \: \textrm{atau}\: \: 8\\ \textrm{d}.&3\: \: \textrm{atau}\: \: -8\\ \textrm{e}.&3\: \: \textrm{atau}\: \: 8 \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{c}\\ &\begin{aligned}\left | 2x-5 \right |&=11\\ (2x-5)&=\pm 11\\ 2x&=5\pm 11\\ 2x&=\begin{cases} 5+11 & =16 \\ \textrm{atau}&\\ 5-11 &=-6 \end{cases}\\ x&=\begin{cases} 8 & \\ &\textrm{atau} \\ -3 & \end{cases} \end{aligned}\end{array}.

\begin{array}{ll}\\ \fbox{10}.&\textrm{Himpunan penyelesaian untuk}\: \: -2\left | x-5 \right |=8\: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \textrm{a}.&\left \{ 0 \right \}\\ \textrm{b}.&\left \{ \: \: \right \}\\ \textrm{c}.&\left \{ 4,5 \right \}\\ \textrm{d}.&\left \{ 1,9 \right \}\\ \textrm{e}.&\left \{ -1,9 \right \} \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{b}\\ &\begin{aligned}-2\left | x-5 \right |&=8\\ \left | x-5 \right |&=-4\: ,\quad \textrm{karena bernilai negatif maka tidak ada harga}\: \: x\: \: \textrm{yang memenuhi}\\\\ \therefore \: \: \: \textbf{HP}&=\left \{ \: \: \right \} \end{aligned}\end{array}.

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Lanjutan 3 (Fungsi Eksponen Dan logaritma)

\begin{array}{ll}\\ 11.&\textrm{Nyatakanlah dengan pangkat positif kemudian sederhankanlah} \end{array}\\  \begin{array}{lllllllll}\\ .\quad\quad&a.&\displaystyle 3^{-4}&d.&\displaystyle \left ( \frac{c}{ab} \right )^{-1}&g.&\displaystyle \left ( x+y \right )^{-1}&j.&\displaystyle \frac{a^{-1}+b^{-1}}{a+b}\\ &b.&\left ( a^{-3}b \right )^{-5}&e.&\displaystyle \left ( \frac{3a^{-2}b^{-3}}{a^{-5}b} \right )^{-2}&h.&\displaystyle \frac{1+a^{-1}}{1+a}&k.&\displaystyle \left ( \frac{1}{1+\displaystyle \frac{1}{a}} \right ):\left ( \frac{1}{1+a} \right )\\ &c.&\displaystyle \frac{3^{0}a^{-6}b^{-3}}{a^{5}b^{-1}}&f.&a^{-1}+b^{-1}&i.&\left ( a+b \right )^{-1}\end{array}.

.\quad\: \, \begin{aligned}&\textrm{Jawab}:\\ &\begin{array}{|c|l|c|l|c|l|c|l|}\hline \multicolumn{2}{|c|}{\textrm{Pembahasan}}&\multicolumn{2}{|c|}{\textrm{Pembahasan}}&\multicolumn{2}{|c|}{\textrm{Pembahasan}}&\multicolumn{2}{|c|}{\textrm{Pembahasan}}\\\hline a&3^{-4}=\displaystyle \frac{1}{3^{4}}&d&\displaystyle \left ( \frac{c}{ab} \right )^{-1}=\left ( \frac{ab}{c} \right )&g&\left ( x+y \right )^{-1}=\displaystyle \frac{1}{\left ( x+y \right )}&j&\begin{aligned}\displaystyle \frac{a^{-1}+b^{-1}}{a+b}&\\ =\displaystyle \frac{\frac{1}{a}+\frac{1}{b}}{a+b}&\\ =\displaystyle \frac{\left ( \frac{a+b}{ab} \right )}{a+b}&\\ =\displaystyle \frac{1}{ab} \end{aligned}\\\hline f.&\multicolumn{3}{|c|}{\begin{aligned}a^{-1}+b^{-1}&=\displaystyle \frac{1}{a}+\frac{1}{b}\\ &=\displaystyle \frac{a+b}{ab}\\ & \end{aligned}}&k.&\multicolumn{3}{|c|}{\begin{aligned}\left ( \displaystyle \frac{1}{1+\displaystyle \frac{1}{a}} \right ):\left ( \displaystyle \frac{1}{1+a} \right )&=\displaystyle \frac{1+a}{1+\displaystyle \frac{1}{a}}\\ &=\displaystyle \frac{1+a}{\left ( \displaystyle \frac{a+1}{a} \right )}\\ &=a \end{aligned}}\\\hline c&\multicolumn{7}{|c|}{\begin{aligned}\displaystyle \frac{3^{0}a^{-6}b^{-3}}{a^{5}b^{-1}}&=\displaystyle \frac{3^{0}}{a^{5}.a^{6}.b^{-1}.b^{3}}\\ &=\displaystyle \frac{1}{a^{11}b^{2}} \end{aligned}}\\\hline \end{array} \end{aligned}.

\begin{array}{ll}\\ 12.&\textrm{Nyatakanlah dengan pangkat positif kemudian sederhankanlah} \end{array}\\ \begin{array}{lllllllll}\\ .\quad\quad&a.&\displaystyle \frac{x^{-2}+y^{-2}}{x^{-1}+y^{-1}}&b.&\displaystyle \frac{x^{-2}y-xy^{-2}}{x^{-1}y-xy^{-1}}&c.&\displaystyle \frac{1-x^{-2}y^{2}}{1-x^{-1}y}\\ &&&&&&&\\ &d.&\displaystyle \frac{3x^{-1}-y^{-2}}{x^{-2}+2y^{-1}}&e.&\displaystyle \frac{x^{-1}y-xy^{-1}}{x^{-1}-y^{-1}}\end{array}.

.\quad\: \, \begin{aligned}&\textrm{Jawab}:\\ &\begin{array}{|c|l|c|l|}\hline \multicolumn{2}{|c|}{\textrm{Pembahasan}}&\multicolumn{2}{|c|}{\textrm{Pembahasan}}\\\hline 12.a&\begin{aligned}\displaystyle \frac{x^{-2}+y^{-2}}{x^{-1}+y^{-1}}&=\frac{\displaystyle \frac{1}{x^{2}}+\frac{1}{y^{2}}}{\displaystyle \frac{1}{x}+\frac{1}{y}}\times \displaystyle \frac{x^{2}y^{2}}{x^{2}y^{2}}\\ &=\frac{\displaystyle y^{2}+x^{2}}{\displaystyle xy^{2}+x^{2}y}\\ &=\displaystyle \frac{\left ( x+y \right )^{2}-2xy}{xy\left ( x+y \right )}\\ &=\displaystyle \frac{x+y}{xy}-\frac{2}{x+y} \end{aligned}&12.c&\begin{aligned}\displaystyle \frac{1-x^{-2}y^{2}}{1-x^{-1}y}&=\frac{1-\displaystyle \frac{y^{2}}{x^{2}}}{1-\displaystyle \frac{y}{x}}\times \displaystyle \frac{x^{2}}{x^{2}}\\ &=\displaystyle \frac{x^{2}-y^{2}}{x^{2}-xy}\\ &=\displaystyle \frac{\left ( x+y \right )\left ( x-y \right )}{x\left ( x-y \right )}\\ &=\displaystyle \frac{x+y}{x} \end{aligned}\\\hline \end{array} \end{aligned}.

\begin{array}{ll}\\ 13.&\textrm{sederhankanlah bilangan di bawah tanda akar berikut ini!}\\ &\begin{array}{lllllllll}\\ a.&\displaystyle \sqrt{20}&b.&\sqrt{48}&c.&\sqrt{96}\\ &&&&&&&\\ d.&\sqrt{120}&e.&\sqrt{200}&f.&\displaystyle \frac{2}{5}\sqrt{500}\end{array}\\\\ &\textrm{Jawab}:\\ &\begin{array}{|l|l|l|l|l|l|}\hline a.&\sqrt{20}=\sqrt{4.5}=\sqrt{4}.\sqrt{5}=2\sqrt{5}&b.&\sqrt{48}=\sqrt{16.3}=4\sqrt{3}&c.&\sqrt{96}=\sqrt{16.6}=4\sqrt{6}\\\hline d.&\sqrt{120}=\sqrt{4.30}=2\sqrt{30}&e.&\sqrt{200}=\sqrt{100.2}=10\sqrt{2}&f.&\begin{aligned}\displaystyle &\frac{2}{5}\sqrt{500}=\displaystyle \frac{2}{5}\sqrt{100.5}\\ &=\displaystyle \frac{2}{5}.10\sqrt{5}=4\sqrt{5} \end{aligned}\\\hline \end{array} \end{array}.

\begin{array}{ll}\\ 14.&\textrm{sederhankanlah bentuk akar berikut ini jika terdefinisi}\\ &\begin{array}{lllllllll}\\ a.&\displaystyle \sqrt{x^{5}}&b.&\sqrt{9a^{2}b}&c.&\sqrt{18a^{3}}\\ &&&&&&&\\ d.&\sqrt{49a^{4}b}&e.&\sqrt{72c^{6}d^{10}}&f.&\displaystyle \sqrt{28p^{8}q^{6}r^{4}}\end{array}\\\\ &\textrm{Jawab}:\\ &\begin{array}{|l|l|l|l|l|l|}\hline a.&\begin{aligned}\sqrt{x^{5}}&=\sqrt{x^{4}.x^{1}}=x^{\frac{4}{2}}.x^{\frac{1}{2}}\\ &=x^{2}.x^{\frac{1}{2}}=x\sqrt{x} \end{aligned}&b.&\sqrt{9a^{2}b}=\sqrt{3^{2}.a^{2}.b}=3a\sqrt{b}&c.&\begin{aligned}\sqrt{18a^{3}}&=\sqrt{9.2.a^{2}.a}\\ &=3a\sqrt{2a} \end{aligned}\\\hline d.&\begin{aligned}\sqrt{49a^{4}b}&=\sqrt{7^{2}.\left ( a^{2} \right )^{2}b}\\ &=7a^{2}\sqrt{b} \end{aligned}&e.&\begin{aligned}\sqrt{72c^{6}d^{10}}&=\sqrt{36.2.\left ( c^{3} \right )^{2}\left ( d^{5} \right )^{2}}\\ &=6c^{3}d^{5}\sqrt{2} \end{aligned}&f.&\begin{aligned}\displaystyle &\sqrt{28p^{8}q^{6}r^{4}}\\ &=\sqrt{4.7\left ( p^{4} \right )^{2}\left ( q^{3} \right )^{2}\left ( r^{2} \right )^{2}}\\ &=2p^{4}q^{3}r^{2}\sqrt{7}\end{aligned}\\\hline \end{array} \end{array}.

\begin{array}{ll}\\ 15.&\textrm{sederhankanlah bentuk akar berikut ini}\\ &\begin{array}{lllllllll}\\ a.&\displaystyle 2\sqrt{3}+4\sqrt{3}&b.&8\sqrt{3}+4\sqrt{3}+\sqrt{3}&c.&4\sqrt{5}+3\sqrt{500}\\ &&&&&&&\\ d.&8\sqrt{5}-2\sqrt{5}&e.&6\sqrt{5}-3\sqrt{5}-\sqrt{5}&f.&\displaystyle 3\sqrt{45}-\sqrt{80}\\ &&&&&&&\\ g.&2\sqrt{3}+3\sqrt{4}+4\sqrt{5}&h.&\displaystyle \sqrt{3}-\frac{1}{2}\sqrt{3}-\frac{1}{3}\sqrt{3}&i.&3\sqrt{24}+2\sqrt{72}-\sqrt{108}\\&&&&&&&\\ j.&\sqrt{2}+\sqrt{8}+\sqrt{32}+\sqrt{128}&k.&\sqrt{8}-\sqrt{20}+3\sqrt{45}-\sqrt{72}&l.&\sqrt{52}-2\sqrt{13}+\sqrt{1024}+\sqrt{32}\end{array} \end{array}.

.\quad\: \, \begin{aligned}&\textrm{Jawab}:\\ &\begin{array}{|c|l|c|l|c|l|c|l|}\hline \multicolumn{2}{|c|}{\textrm{Pembahasan}}&\multicolumn{2}{|c|}{\textrm{Pembahasan}}&\multicolumn{2}{|c|}{\textrm{Pembahasan}}&\multicolumn{2}{|c|}{\textrm{Pembahasan}}\\\hline a.&\begin{aligned}2\sqrt{3}+4\sqrt{3}&=\left ( 2+4 \right )\sqrt{3}\\ &=6\sqrt{3} \end{aligned}&b.&\begin{aligned}8\sqrt{3}&+4\sqrt{3}+\sqrt{3}\\ &=\left ( 8+4+1 \right )\sqrt{3}\\ &=13\sqrt{3} \end{aligned}&c.&\begin{aligned}4\sqrt{5}+3\sqrt{500}&=4\sqrt{5}+3\sqrt{100\times 5}\\ &=4\sqrt{5}+3\left ( 10\sqrt{5} \right )\\ &=4\sqrt{5}+30\sqrt{5}\\ &=\left ( 4+30 \right )\sqrt{5}\\ &=34\sqrt{5} \end{aligned}&d.&\begin{aligned}8\sqrt{5}-2\sqrt{5}=6\sqrt{5} \end{aligned}\\\hline k.&\multicolumn{3}{|l|}{\begin{aligned}\sqrt{8}-\sqrt{20}+&3\sqrt{45}-\sqrt{72}\\ &=\sqrt{4.2}-\sqrt{4.5}+3\sqrt{9.5}-\sqrt{36.2}\\ &=2\sqrt{2}-2\sqrt{5}+3.3\sqrt{5}-6\sqrt{2}\\ &=\left (2-6 \right )\sqrt{2}+\left ( 9-2 \right )\sqrt{5}\\ &=-4\sqrt{2}+7\sqrt{5} \end{aligned}}&l.&\multicolumn{3}{|l|}{\begin{aligned}\sqrt{52}-2\sqrt{13}+&\sqrt{1024}+\sqrt{32}\\ &=\sqrt{4.13}-2\sqrt{13}+\sqrt{32\times 32}+\sqrt{16.2}\\ &=2\sqrt{13}-2\sqrt{13}+32-4\sqrt{2}\\ &=0+32-4\sqrt{2}\\ &=32-4\sqrt{2} \end{aligned}}\\\hline \end{array} \end{aligned}.

\begin{array}{ll}\\ 16.&\textrm{sederhankanlah bentuk akar berikut ini}\\ &\begin{array}{lllllllll}\\ a.&\displaystyle \sqrt{3}\left ( 7+\sqrt{3} \right )&b.&4\sqrt{2}\left ( 3\sqrt{14}-\sqrt{18} \right )&c.&\left ( 5+2\sqrt{6} \right )\left ( 5+2\sqrt{6} \right )\\ &&&&&&&\\ d.&\left ( 7-\sqrt{11} \right )\left ( 7-\sqrt{11} \right )&e.&\left ( 5\sqrt{3}+2\sqrt{7} \right )^{2}&f.&\left ( 5\sqrt{2}-6 \right )^{2} \end{array}\\\\ &\textrm{Jawab}:\\ &\begin{array}{|c|l|c|l|c|l|c|l|}\hline \multicolumn{2}{|c|}{\textrm{Pembahasan}}&\multicolumn{2}{|c|}{\textrm{Pembahasan}}&\multicolumn{2}{|c|}{\textrm{Pembahasan}}&\multicolumn{2}{|c|}{\textrm{Pembahasan}}\\\hline a.&\begin{aligned}&\sqrt{3}\left ( \sqrt{7}+\sqrt{3} \right )\\ &=\sqrt{3}\sqrt{7}+\sqrt{3}.\sqrt{3}\\ &=\sqrt{3.7}+\sqrt{3^{2}}\\ &=\sqrt{21}+3 \end{aligned}&b.&\begin{aligned}&4\sqrt{2}\left ( 3\sqrt{14}-\sqrt{18} \right )\\ &=4.3.\sqrt{2.14}-4\sqrt{2.18}\\ &=12\sqrt{2.2.7}-4\sqrt{2.2.9}\\ &=12.2\sqrt{7}-4.2.3\\ &=24\sqrt{7}-24\end{aligned}&c.&\begin{aligned}&\left ( 5+2\sqrt{6} \right )\left ( 5+2\sqrt{6} \right )\\ &=\left ( 5+2\sqrt{6} \right )^{2}\\ &\textrm{ingat bentuk}\\ &\left ( a+b \right )^{2}=a^{2}+2ab+b^{2}\\ &=5^{2}+2\left ( 5.2\sqrt{6} \right )+\left ( 2\sqrt{6} \right )^{2}\\ &=25+20\sqrt{6}+4.6\\ &=25+24+20\sqrt{6}\\ &=49+20\sqrt{6} \end{aligned}&d.&\begin{aligned}&\left ( 7-\sqrt{11} \right )\left ( 7-\sqrt{11} \right )\\ &=.............\\ &=.............\\ &=.............\\ &=.............\\ &=.............\\ &=....... \end{aligned}\\\hline \end{array} \end{array}.

\begin{array}{ll}\\ 17.&\textrm{sederhankanlah bentuk akar berikut ini}\\ &\begin{array}{lllllllll}\\ a.&\sqrt{8+2\sqrt{15}}&b.&\sqrt{3-2\sqrt{2}}&c.&\sqrt{9+6\sqrt{2}}\\ &&&&&&&\\ d.&\sqrt{9-4\sqrt{5}}&e.&\sqrt{7+\sqrt{40}}&f.&\sqrt{\displaystyle \frac{5}{2}+\sqrt{6}}\\ \end{array}\\ \end{array}.

.\quad\: \, \begin{aligned}&\textrm{Jawab}:\\ &\begin{array}{|c|l|c|l|c|l|}\hline \multicolumn{2}{|c|}{\textrm{Pembahasan}}&\multicolumn{2}{|c|}{\textrm{Pembahasan}}&\multicolumn{2}{|c|}{\textrm{Pembahasan}}\\\hline a.&\begin{aligned}\sqrt{8+2\sqrt{15}}&=\sqrt{5+3+2\sqrt{5.3}}\\ &=\sqrt{\left ( 5 \right )^{2}+\left ( 3 \right )^{2}+2\sqrt{5.3}}\\ &=\sqrt{\left ( \sqrt{5}+\sqrt{3} \right )^{2}}\\ &=\sqrt{5}+\sqrt{3} \end{aligned}&b.&\begin{aligned}\sqrt{3-2\sqrt{2}}&=\sqrt{2+1-2\sqrt{2.1}}\\ &=\sqrt{\left ( 2 \right )^{2}+1^{2}-2\sqrt{2.1}}\\ &=\sqrt{\left ( \sqrt{2}-1 \right )^{2}}\\ &=\sqrt{2}-1 \end{aligned}&c.&\begin{aligned}\sqrt{9+6\sqrt{2}}&=\sqrt{9+2.3\sqrt{2}}\\ &=\sqrt{9+2\sqrt{9}.\sqrt{2}}\\ &=\sqrt{9+2\sqrt{18}}\\ &=\sqrt{6+3+2\sqrt{6.3}}\\ &=\sqrt{\left ( \sqrt{6}+\sqrt{3} \right )^{2}}\\ &=\sqrt{6}+\sqrt{3} \end{aligned}\\\hline d.&\begin{aligned}\sqrt{9-4\sqrt{5}}&=\sqrt{9-2.2\sqrt{5}}\\ &=\sqrt{9-2\sqrt{4}.\sqrt{5}}\\ &=\sqrt{5+4-2\sqrt{5.4}}\\ &=\sqrt{\left ( \sqrt{5}+\sqrt{4} \right )^{2}}\\ &=\sqrt{5}+\sqrt{4}\\ &=\sqrt{5}+2 \end{aligned}&e.&\begin{aligned}\sqrt{7+\sqrt{40}}&=\sqrt{7+\sqrt{4.10}}\\ &=\sqrt{7+2\sqrt{10}}\\ &=\sqrt{5+2+2\sqrt{5.2}}\\ &=\sqrt{\left ( \sqrt{5}+\sqrt{2} \right )^{2}}\\ &=\sqrt{5}+\sqrt{2}\end{aligned}&f.&\begin{aligned}\sqrt{\displaystyle \frac{5}{2}+\sqrt{6}}&=\sqrt{\displaystyle \frac{(3+2)}{2}+2.\frac{1}{2}\sqrt{3.2}}\\ &=\sqrt{\displaystyle \frac{(3+2)}{2}+2\sqrt{\displaystyle \frac{3.2}{4}}}\\ &=\sqrt{\displaystyle \frac{3}{2}+\frac{2}{2}+2\sqrt{\displaystyle \frac{3}{2}.\frac{2}{2}}}\\ &=\sqrt{\left ( \displaystyle \sqrt{\frac{3}{2}}+1 \right )^{2}}\\ &=\sqrt{\displaystyle \frac{3}{2}}+1\\ &=\sqrt{\displaystyle \frac{2.3}{2.2}}+1\\ &=\displaystyle \frac{1}{2}\sqrt{6}+1 \end{aligned} \\\hline \end{array} \end{aligned}.

\begin{array}{ll}\\ 18.&\textrm{Tentukanlah nilai x yang memenuhi persamaan berikut} \end{array}\\ \begin{array}{lllllll}\\ .\quad\quad&a.&\displaystyle 2^{x}=8&j.&\displaystyle 5^{x^{2}-8x+12}=1&s.&\displaystyle \sqrt{8^{2x-1}}-4\sqrt[3]{2^{1-x}}=0\\ &b.&\displaystyle 2^{x+1}=32&k.&\displaystyle 3^{x^{2}+3x}=3^{x+8}&t.&\displaystyle 3^{5-x}=\frac{1}{27}\sqrt[3]{3^{x}}=0\\ &c.&\displaystyle 2^{3x-4}=32&l.&\displaystyle 25^{x^{2}-5x+7}=125^{x-2}&u.&\displaystyle \sqrt[3]{8^{x+2}}=\left ( \frac{1}{32} \right )^{2-x}\\ &d.&\displaystyle 3^{2x-1}=\frac{1}{27}&m.&\displaystyle 2^{x^{2}+x}=4^{x+1}&v.&\displaystyle 4^{1-2x}=\frac{16^{2-x}}{32^{1-x}}\\ &e.&\displaystyle 3^{5x+2}=9^{x+4}&n.&\displaystyle 4^{x+3}=\sqrt[3]{8^{x+5}}&w.&\displaystyle \left ( \frac{1}{2} \right )^{2x+1}=\sqrt{\frac{2^{4x-1}}{128}}\\ &f.&\displaystyle 5^{x-9}=25^{3-x}&o.&\displaystyle 9^{3x+2}=\frac{1}{81^{2x-5}}&x.&\displaystyle \left ( \frac{2}{3} \right )^{2x-3}=\left ( \frac{27}{8} \right )^{3-2x}\\ &g.&\displaystyle 4^{2x-1}=1&p.&\displaystyle \sqrt{2^{x-5}}=2\sqrt{2}&y.&\displaystyle x^{x^{x^{x^{...}}}}=2\\ &h.&\displaystyle 5^{x-1}=\sqrt{5}&q.&\displaystyle \left ( \frac{1}{4} \right )^{x-1}=\sqrt[3]{2^{3x+1}}&z.&\displaystyle 2^{x+5}+2^{5-x}=64\\ &i.&\displaystyle 7^{4x}=\frac{7}{\sqrt[4]{7}}&r.&\displaystyle \frac{2^{x}}{8^{x+2}}=64.4^{x}& \end{array}.

.\quad \: \, \begin{aligned}&\textrm{Jawab}:\\ &\begin{array}{|c|l|c|l|c|l|c|l|c|l|c|l|}\hline \multicolumn{2}{|c|}{\textrm{Pembahasan}}&\multicolumn{2}{|c|}{\textrm{Pembahasan}}&\multicolumn{2}{|c|}{\textrm{Pembahasan}}&\multicolumn{2}{|c|}{\textrm{Pembahasan}}&\multicolumn{2}{|c|}{\textrm{Pembahasan}}&\multicolumn{2}{|c|}{\textrm{Pembahasan}}\\\hline a&\begin{aligned}\displaystyle 2^{x}&=8\\ \displaystyle 2^{x}&=2^{3}\\ x&=3 \end{aligned}&b&\begin{aligned}\displaystyle 2^{x+1}&=32\\ \displaystyle 2^{x+1}&=2^{5}\\ x+1&=5\\ x&=4 \end{aligned}&c&\begin{aligned}\displaystyle 2^{3x-4}&=32\\ \displaystyle 2^{3x-4}&=2^{5}\\ 3x-4&=5\\ 3x&=9\\ x&=3 \end{aligned}&d&\begin{aligned}3^{2x-1}&=\displaystyle \frac{1}{27}\\ \displaystyle 3^{2x-1}&=\displaystyle \frac{1}{3^{3}}\\ \displaystyle 3^{2x-1}&=3^{-3}\\ 2x-1&=-3\\ 2x&=-2\\ x&=-1 \end{aligned}&e&\begin{aligned}\displaystyle 3^{5x+2}&=9^{x+4}\\ \displaystyle 3^{5x+2}&=3^{2\left ( x+4 \right )}\\ 5x+2&=2x+8\\ 5x-2x&=8-2\\ 3x&=6\\ x&=2 \end{aligned}&f&\begin{aligned}\displaystyle 5^{x-9}&=25^{3-x}\\ \displaystyle 5^{x-9}&=5^{2\left ( 3-x \right )}\\ x-9&=6-2x\\ x+2x&=6+9\\ 3x&=15\\ x&=5 \end{aligned}\\\hline \end{array} \end{aligned}.

.\quad \: \, \begin{aligned} &\begin{array}{|c|l|c|l|c|l|c|l|c|l|}\hline \multicolumn{2}{|c|}{\textrm{Pembahasan}}&\multicolumn{2}{|c|}{\textrm{Pembahasan}}&\multicolumn{2}{|c|}{\textrm{Pembahasan}}&\multicolumn{2}{|c|}{\textrm{Pembahasan}}&\multicolumn{2}{|c|}{\textrm{Pembahasan}}\\\hline g&\begin{aligned}\displaystyle 4^{2x-1}&=1\\ \displaystyle 4^{2x-1}&=4^{0}\\ 2x-1&=0\\ 2x&=1\\ x&=\displaystyle \frac{1}{2} \end{aligned}&h&\begin{aligned}\displaystyle 5^{x-1}&=\sqrt{5}\\ \displaystyle 5^{x-1}&=5^{\frac{1}{2}}\\ x-1&=\displaystyle \frac{1}{2}\\ x&=\displaystyle 1\frac{1}{2}\\ x&=\displaystyle \frac{3}{2} \end{aligned}&i&\begin{aligned}\displaystyle 7^{4x}&=\displaystyle \frac{7}{\sqrt[4]{7}}\\ \displaystyle 7^{4x}&=\displaystyle \frac{7^{1}}{7^{\frac{1}{4}}}\\ \displaystyle 7^{4x}&=7^{\left (1-\frac{1}{4} \right )}\\ 4x&=\displaystyle \frac{3}{4}\\ x&=\displaystyle \frac{3}{16}\end{aligned}&j&\begin{aligned}\displaystyle 5^{x^{2}-8x+12}&=1\\ \displaystyle 5^{x^{2}-8x+12}&=5^{0}\\ \displaystyle x^{2}-8x+12&=0\\ \left ( x-2 \right )\left ( x-6 \right )&=0\\ x=2\quad \textrm{atau}\quad x=6& \end{aligned}&k&\begin{aligned}\displaystyle 3^{x^{2}+3x}&=3^{x+8}\\ x^{2}+3x&=x+8\\ x^{2}+3x-x-8&=0\\ x^{2}+2x-8&=0\\ \left ( x+4 \right )\left ( x-2 \right )&=0\\ x=-4\quad \textrm{atau}\quad x=2& \end{aligned}\\\hline \end{array} \end{aligned}.

.\quad \: \, \begin{aligned} &\begin{array}{|c|l|c|l|c|l|}\hline \multicolumn{2}{|c|}{\textrm{Pembahasan}}&\multicolumn{2}{|c|}{\textrm{Pembahasan}}&\multicolumn{2}{|c|}{\textrm{Pembahasan}}\\\hline l&\begin{aligned}\displaystyle 25^{x^{2}-5x+7}&=125^{x-2}\\ \displaystyle \left ( 5^{2} \right )^{x^{2}-5x+7}&=\left ( 5^{3} \right )^{x-2}\\ 2x^{2}-10x+14&=3x-6\\ 2x^{2}-10x-3x+14+6&=0\\ 2x^{2}-13x+20&=0\\ \left ( 2x-5 \right )\left ( x-4 \right )&=0\\ x=\displaystyle \frac{5}{2}\quad \textrm{atau}\quad x=4&\end{aligned}&m&\begin{aligned}\displaystyle 2^{x^{2}+x}&=4^{x+1}\\ 2^{x^{2}+x}&=\left ( 2^{2} \right )^{x+1}\\ x^{2}+x&=2x+2\\ x^{2}+x-2x-2&=0\\ x^{2}-x-2&=0\\ \left ( x-2 \right )\left ( x+1 \right )&=0\\ x=2\quad \textrm{atau}\quad x=-1&\end{aligned}&n&\begin{aligned}\displaystyle 4^{x+3}&=\sqrt[3]{8^{x+5}}\\ \displaystyle \left ( 2^{2} \right )^{x+3}&=\left ( 2^{3} \right )^{\frac{x+5}{3}}\\ 2x+6&=x+5\\ x&=-1\end{aligned}\\\hline \end{array} \end{aligned}.

.\quad \: \, \begin{aligned} &\begin{array}{|c|l|c|l|c|l|c|l|}\hline \multicolumn{2}{|c|}{\textrm{Pembahasan}}&\multicolumn{2}{|c|}{\textrm{Pembahasan}}&\multicolumn{2}{|c|}{\textrm{Pembahasan}}&\multicolumn{2}{|c|}{\textrm{Pembahasan}}\\\hline w&\begin{aligned}\displaystyle \left ( \frac{1}{2} \right )^{2x+1}&=\sqrt{\frac{2^{4x-1}}{128}}\\ \left ( 2^{-1} \right )^{2x+1}&=\left ( 2^{4x-1-7} \right )^{\frac{1}{2}}\\ -2x-1&=2x-4\\ -4x&=-3\\ x&=\displaystyle \frac{3}{4}\end{aligned}&x&\begin{aligned}\displaystyle \left ( \frac{2}{3} \right )^{2x-3}&=\left ( \frac{27}{8} \right )^{3-2x}\\ \displaystyle \left ( \frac{2}{3} \right )^{2x-3}&=\left ( \left ( \frac{2}{3} \right )^{-3} \right )^{3-2x}\\ 2x-3&=-9+6x\\ -4x&=-6\\ x&=\displaystyle \frac{3}{2} \end{aligned}&y&\begin{aligned}\displaystyle x^{x^{x^{x^{x^{...}}}}} &=2\\ \displaystyle x^{\underset{2}{\underbrace{x^{x^{x^{...}}}}}}&=2\\ \displaystyle x^{2}&=2\\ x&=\sqrt{2}\end{aligned}&z&\begin{aligned}\displaystyle 2^{x+5}+2^{5-x}&=64\\ 2^{x}.2^{5}+2^{5}.2^{-x}&=64\\ 32.2^{x}+\displaystyle \frac{32}{2^{x}}&=64\\ 2^{x}+\displaystyle \frac{1}{2^{x}}&=2\\ \left ( 2^{x} \right )^{2}+1&=2\left ( 2^{x} \right )\\ \left ( 2^{x} \right )^{2}-2\left ( 2^{x} \right )+1&=0\\ \left ( 2^{x}-1 \right )\left ( 2^{x}-1 \right )&=0\\ \left ( 2^{x}-1 \right )^{2}&=0\\ 2^{x}-1&=0\\ 2^{x}&=1\\ 2^{x}&=2^{0}\\ x&=0\end{aligned}\\\hline \end{array} \end{aligned}.

\begin{array}{ll}\\ 19&\textrm{Carilah nilai x dan y yang memenuhi persamaan berikut}\\ &a.\quad \begin{cases} 3^{x-2y} & =\displaystyle \frac{1}{81} \\ 2^{x-y} & = 16 \end{cases}\\ &b.\quad \begin{cases} 5^{x-2y+1} & =25^{x-2y} \\ 4^{x-y+2} & =32^{x-2y+1} \end{cases}\\ &c.\quad \begin{cases} 4^{x-2y+1} & =8^{2x-y} \\ 3^{x+y+1} &= 9^{2x-y-4} \end{cases} \end{array}.

.\quad \: \, \begin{aligned}&\textrm{Jawab}:\\ &\textrm{a. Diketahui bahwa}\\ & \quad \begin{cases} 3^{x-2y} \\ & =\displaystyle \frac{1}{81} \\ 2^{x-y} & = 16 \end{cases}\\ &\begin{array}{l|l|l}\\ \begin{aligned}(\ast )\qquad 3^{x-2y}&=\displaystyle \frac{1}{81}\\ 3^{x-2y}&=3^{-4}\\ x-2y&=-4\\ (\ast \ast )\qquad 2^{x-y}&=16\\ 2^{x-y}&=2^{4}\\ x-y&=4 \end{aligned}&\qquad \begin{matrix} x-2y=-4\: \: ..............(\ast )\\ x-y=4\: \: .................(\ast \ast ) \end{matrix}&\textrm{dengan eliminasi kita akan mendapatkan}\begin{cases} & x=12 \\ & y=8 \end{cases}\\ && \end{array} \end{aligned}.

\begin{array}{lll}\\ 20.&\textrm{Sederhanakanlah}\\ &&\\ &a.\quad \displaystyle \frac{3^{m+n}\times 3^{n+1}\times 3^{n-m}}{3^{2n+1}\times 3^{n+1}}&c.\quad \displaystyle \frac{\left ( 2^{m+2} \right )^{2}-2^{2}.2^{2m}}{2^{m}.2^{m+2}}\\ &&\\ &b.\quad \displaystyle \frac{3^{n+1}-3^{n}}{3^{n}+3^{n-1}}\\\\ &\textrm{Jawab}:\\\\ &\begin{aligned}a.\quad \displaystyle \frac{3^{m+n}\times 3^{n+1}\times 3^{n-m}}{3^{2n+1}\times 3^{n+1}}&=\displaystyle \frac{3^{m}.3^{n}\times 3^{n}.3\times 3^{n}.3^{-m}}{3^{2n}.3\times 3^{n}.3}\\ &=\displaystyle \frac{3^{m+n+n+1+n-m}}{3^{2n+1+n+1}}\\ &=\displaystyle \frac{3^{3n+1}}{3^{3n+2}}=\frac{3^{3n}.3}{3^{3n}.3^{2}}\\ &=\displaystyle \frac{3}{9}\\ &=\displaystyle \frac{1}{3}\end{aligned} \\ &\begin{aligned}b.\quad \displaystyle \frac{3^{n+1}-3^{n}}{3^{n}+3^{n-1}}&=\displaystyle \frac{3^{n}.3-3^{n}}{3^{n}+3^{n}.3^{-1}}\\ &=\displaystyle \frac{3^{n}\left ( 3-1 \right )}{3^{n}\left ( 1-3^{-1} \right )}=\displaystyle \frac{3^{n}\left ( 3-1 \right )}{3^{n}\left ( 1-\frac{1}{3} \right )}\\ &=\displaystyle \frac{2}{\frac{2}{3}}\\ &=3\end{aligned} \end{array}.

Sumber Referensi

  1. Kurnianingsih, Sri, Kuntarti & Sulistiyono. 2007. Matematika SMA dan MA Untuk Kelas X Semester 1. Jakarta: ESIS.
  2. Noormandiri, Endar Sucipto. 2000. Buku Pelajaran Matematika SMU untuk Jilid 1 Kelas 1 Catur Wulan 1, 2, dan 3. Jakarta: Erlangga.
  3. Sunardi, Slamet Waluyo, Sutrisno & Subagya. 2004. Matematika 1A untuk SMA Kelas 1. Jakarta: Bumi Aksara.
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Lanjutan 2 (Fungsi Eksponen dan Logaritma)

\begin{array}{ll}\\ 7.&\textrm{Hitunglah} \end{array}\\ \begin{array}{lllllllll}\\ .\quad\quad&a.&\displaystyle \frac{ \sqrt{1+\sqrt{1+\sqrt{1+...}}}}{\sqrt{1-\sqrt{1-\sqrt{1-...}}}}&b.&\displaystyle \frac{ \sqrt{2+\sqrt{2+\sqrt{2+...}}}}{\sqrt{2-\sqrt{2-\sqrt{2-...}}}}\\ &&&&&&&\\ &c.&\displaystyle \frac{\sqrt{31-\sqrt{31-\sqrt{31-...}}}}{\sqrt{1+\sqrt{1+\sqrt{1+...}}}}&d.&\displaystyle \frac{\sqrt{31+\sqrt{31+\sqrt{31+...}}}}{\sqrt{3+\sqrt{3+\sqrt{3+...}}}}\\ \end{array}..

.\quad\: \, \begin{aligned}&\textrm{Jawab}:\\ &\begin{array}{|l|l|}\hline \multicolumn{2}{|c|}{\textbf{Pembahasan}}\\\hline \begin{aligned}x&=\sqrt{1+\sqrt{1+\sqrt{1+...}}}\\ x^{2}&=1+\underset{x}{\underbrace{\sqrt{x+\sqrt{x+...}}}}\\ x^{2}&=1+x,\quad \textrm{gunakan rumus \textbf{abc}}\\ x^{2}&-x-1=0\begin{cases} & a=1 \\ & b=-1 \\ & c=-1 \end{cases}\\ x_{1,2}&=\displaystyle \frac{-b\pm \sqrt{b^{2}-4ac}}{2a} \end{aligned}&\begin{aligned}x&=\sqrt{1-\sqrt{1-\sqrt{1-...}}}\\ x^{2}&=1-\underset{x}{\underbrace{\sqrt{x-\sqrt{x-...}}}}\\ x^{2}&=1-x,\quad \textrm{gunakan rumus \textbf{abc}}\\ x^{2}&+x-1=0\begin{cases} & a=1 \\ & b=1 \\ & c=-1 \end{cases}\\ x_{1,2}&=\displaystyle \frac{-b\pm \sqrt{b^{2}-4ac}}{2a} \end{aligned}\\\hline \multicolumn{2}{|c|}{\textrm{kita ambil akar yang positif saja}}\\\hline x=\displaystyle \frac{1+\sqrt{5}}{2}&x=\displaystyle \frac{-1+\sqrt{5}}{2}\\\hline \multicolumn{2}{|c|}{\begin{aligned}\displaystyle \frac{ \sqrt{1+\sqrt{1+\sqrt{1+...}}}}{\sqrt{1-\sqrt{1-\sqrt{1-...}}}}&=\displaystyle \frac{\displaystyle \frac{1+\sqrt{5}}{2}}{\displaystyle \frac{-1+\sqrt{5}}{2}}\\ &=\displaystyle \frac{\sqrt{5}+1}{\sqrt{5}-1}\\ &=\displaystyle \frac{\sqrt{5}+1}{\sqrt{5}-1}\times \displaystyle \frac{\sqrt{5}+1}{\sqrt{5}+1}\\ &=\displaystyle \frac{\sqrt{25}+2\sqrt{5}+1}{\sqrt{25}-1}\\ &=\displaystyle \frac{5+2\sqrt{5}+1}{5-1}\\ &=\displaystyle \frac{6+2\sqrt{5}}{4}\\ &=\displaystyle \frac{1}{2}\left ( 3+\sqrt{5} \right ) \end{aligned}}\\\hline \end{array} \end{aligned}.

\begin{array}{lp{15.0cm}}\\ 8.&\textrm{Perhatikanlah bentuk akar berikut yang dikemukankan oleh matematikawan India \textbf{Srinivasa Ramanujan} yang terkenal dengan istilah \textbf{Ramanujan's infinite radicals}, yaitu} \end{array}\\ \begin{array}{lll}\\ .\quad\quad&.&\displaystyle \sqrt{1+2\sqrt{1+3\sqrt{1+4\sqrt{1+5\sqrt{1+6\sqrt{1+...}}}}}}\\ &&\\ &&\textrm{Tentukanlah nilai akar tersebut}? \end{array}.

.\quad\: \, \begin{aligned}&\textrm{Jawab}:\\ &\textrm{Untuk menemukan solusi soal di atas, saat} \quad x>0,\: \: \textrm{yaitu}:\\ &\begin{aligned}x^{2}&=x^{2}\\ x^{2}&=1+\left ( x^{2}-1 \right )\\ &=1+\left ( x-1 \right )\left ( x+1 \right )\\ &=1+\left ( x-1 \right )\sqrt{\left ( x+1 \right )^{2}}\\ &=1+\left ( x-1 \right )\sqrt{1+\left ( x+1 \right )^{2}-1}\\ &=1+\left ( x-1 \right )\sqrt{1+\left ( x+1-1 \right )\left ( x+1+1 \right )}\\ &=1+\left ( x-1 \right )\sqrt{1+x\left ( x+2 \right )}\\ &=1+\left ( x-1 \right )\sqrt{1+x\sqrt{\left ( x+2 \right )^{2}}}\\ &=1+\left ( x-1 \right )\sqrt{1+x\sqrt{1+\left ( x+2 \right )^{2}-1}}\\ &=1+\left ( x-1 \right )\sqrt{1+x\sqrt{1+\left ( x+2-1 \right )\left ( x+2+1 \right )}}\\ &=1+\left ( x-1 \right )\sqrt{1+x\sqrt{1+\left ( x+1 \right )\left ( x+3 \right )}}\\ &=1+\left ( x-1 \right )\sqrt{1+x\sqrt{1+\left ( x+1 \right )\sqrt{\left ( x+3 \right )^{2}}}}\\ x^{2}&=1+\left ( x-1 \right )\sqrt{1+x\sqrt{1+\left ( x+1 \right )\sqrt{...}}}\\ x&=\sqrt{1+\left ( x-1 \right )\sqrt{1+x\sqrt{1+\left ( x+1 \right )\sqrt{...}}}} \end{aligned} \end{aligned}.

.\quad \: \: \textrm{Jadi},\quad x=3.

\begin{array}{ll}\\ 9.&\textrm{Nyatakan dalam bentuk pangkat} \end{array}\\ \begin{array}{lllllllll}\\ .\quad\quad&a.&\displaystyle \sqrt{6}&b.&\displaystyle 6\sqrt{6}&c.&\displaystyle \sqrt{\sqrt{6\sqrt[3]{6}}}&d.&\displaystyle \sqrt{6\sqrt{6\sqrt{6\sqrt{6}}}}\\ &&&&&&&\\ &e.&\displaystyle \sqrt{\sqrt{\sqrt{\sqrt{6}}}}&f.&\displaystyle \sqrt[4]{\sqrt{\sqrt[3]{\sqrt{6}}}}&g.&\displaystyle \sqrt[3]{\sqrt{a^{2}+2ab+b^{2}}}&h.&\displaystyle \sqrt[5]{x\sqrt{x\sqrt{x\sqrt{x}}}}\left ( \sqrt{x\sqrt[3]{x}}+\sqrt[3]{x\sqrt{x}} \right )\\ &&&&&&&\\ &i.&\displaystyle \left ( \sqrt[3]{16\sqrt[3]{8\sqrt{4}}} \right )^{\frac{1}{4}}&j.&\displaystyle \sqrt[3]{\sqrt[5]{1024a^{15}b^{30}}}&k.&\displaystyle \sqrt[a]{16^{\frac{1}{2}a(a-1)}x^{2a^{2}-a}}&l.&\displaystyle \frac{\sqrt[4]{\left ( a^{4}b^{\frac{2}{3}} \right )^{2}}}{\sqrt[3]{\left ( a^{\frac{1}{2}}b^{3} \right )^{\frac{1}{2}}}}\times \displaystyle \frac{\sqrt[4]{\left ( a^{2\frac{1}{3}}b \right )^{2}}}{\sqrt[3]{a^{\frac{1}{4}}\sqrt[3]{b^{\frac{2}{3}}}}}\\ \end{array}.

.\quad\: \, \begin{aligned}&\textrm{Jawab}:\\ &\begin{array}{|c|l|c|l|c|l|c|l|}\hline \multicolumn{2}{|c|}{\textrm{Pembahasan}}&\multicolumn{2}{|c|}{\textrm{Pembahasan}}&\multicolumn{2}{|c|}{\textrm{Pembahasan}}&\multicolumn{2}{|c|}{\textrm{Pembahasan}}\\\hline a.&\sqrt{6}=6^{\frac{1}{2}}&b.&\begin{aligned}\displaystyle 6\sqrt{6}&=\displaystyle 6^{1}\times 6^{\frac{1}{2}}\\ &=\displaystyle 6^{1+\frac{1}{2}}\\ &=\displaystyle 6^{1\frac{1}{2}}=6^{\frac{3}{2}} \end{aligned}&c.&\begin{aligned}\displaystyle \sqrt[2\times 2]{6\times 6^{\frac{1}{3}}}&=\displaystyle \sqrt[4]{6^{1\frac{1}{3}}}\\ &=\sqrt[4]{6^{\frac{4}{3}}}\\ &=\displaystyle 6^{\frac{4}{3}\times \frac{1}{4}}\\ &=\displaystyle 6^{\frac{1}{3}} \end{aligned}&d.&\begin{aligned}\displaystyle \sqrt{6\sqrt{6\sqrt{6\sqrt{6}}}}&=\displaystyle \left ( 6\left ( 6\left ( 6\left ( 6 \right )^{\frac{1}{2}} \right )^{\frac{1}{2}} \right )^{\frac{1}{2}} \right )^{\frac{1}{2}}\\ &=\displaystyle 6^{\frac{1}{2}}\times 6^{\frac{1}{4}}\times 6^{\frac{1}{8}}\times 6^{\frac{1}{16}}\\ &=\displaystyle 6^{\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}}\\ &=\displaystyle 6^{\frac{8+4+2+1}{16}}\\ &=\displaystyle 6^{\frac{15}{16}}\end{aligned} \\\hline \end{array} \end{aligned}.

\begin{array}{ll}\\ 10.&\textrm{Hitunglah}\:\\\\ &\quad\quad\qquad \displaystyle \sqrt[8]{2207-\displaystyle \frac{1}{2207-\displaystyle \frac{1}{2207-\displaystyle \frac{1}{2207-\displaystyle \frac{1}{...}}}}}\\\\ &\textrm{nyatakan jawabannya dalam bentuk }\: \displaystyle \frac{a\pm b\sqrt{c}}{d}\: \textrm{dengan a, b, c, dan d bilangan-bilangan bulat}\\ \end{array}.

.\quad\: \, \begin{aligned}&\textrm{Jawab}: \end{aligned}.

.\qquad\: \, \textbf{Misalkan}\: \:x=\: \displaystyle \sqrt[8]{2207-\displaystyle \frac{1}{2207-\displaystyle \frac{1}{2207-\displaystyle \frac{1}{2207-\displaystyle \frac{1}{...}}}}} \\\\\\ \qquad\qquad \begin{array}{l|l}\\ \begin{aligned}x^{8}&=2207-\displaystyle \underset{x^{8}}{\underbrace{\displaystyle \frac{1}{2207-\frac{1}{2207-\frac{1}{2207-...}}}}}\\ x^{8}&=2207-\displaystyle \frac{1}{x^{8}}\\ x^{8}+\displaystyle \frac{1}{x^{8}}&=2207\\ \left ( x^{4}+\displaystyle \frac{1}{x^{4}} \right )^{2}&=2207+2\\ \left ( x^{4}+\displaystyle \frac{1}{x^{4}} \right )&=\sqrt{2209}=47 \end{aligned}&\begin{aligned}x^{4}+\displaystyle \frac{1}{x^{4}}&=47\\ \left ( x^{2}+\displaystyle \frac{1}{x^{2}} \right )^{2}&=47+2\\ x^{2}+\displaystyle \frac{1}{x^{2}}&=\sqrt{49}=7\\ \left ( x+\displaystyle \frac{1}{x} \right )^{2}&=7+2\\ x+\displaystyle \frac{1}{x}&=\sqrt{9}=3\\ x^{2}-3x+1&=0,\quad \textrm{persamaan kuadrat dalam x, gunakan rumus abc}\\ x_{1,2}=&\displaystyle \frac{3\pm \sqrt{5}}{2}=\displaystyle \frac{3\pm 1\sqrt{5}}{2}=\displaystyle \frac{a\pm b\sqrt{c}}{d}\\ &\textbf{Sehingga},\quad \begin{cases} & a=3 \\ & b=1 \\ & c=5 \\ & d=2 \end{cases} \end{aligned} \end{array}.

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Lanjutan (Fungsi Eksponen Dan Logaritma)

C. Bilangan Bentuk Akar dan Pangkat Pecahan

\begin{aligned}&\textrm{Jika}\: \: p\: \: \textrm{dan}\: \: q\: \: \textrm{bilangan real dan}\: \: m\: \: \textrm{bilangan bulat positif, maka}\\ &\qquad\qquad\quad \LARGE\boxed{\textbf{p}^{\textbf{n}}=\textbf{q} \quad\Leftrightarrow\quad \sqrt[\textbf{n}]{\textbf{q}}=\textbf{p}}\\ &\textrm{dengan}\\ &\quad \sqrt[n]{q}\: \: \textrm{disebut sebagai akar (radikal)}\\ &\qquad q\: \: \textrm{disebut radikan (bilangan yang ditarik akarnya)}\\ &\qquad n\: \: \textrm{adalah indeks (pangakt dari akar)} \end{aligned}.

Selanjutnya perhatikanlah tabel berikut

\begin{array}{|c|l|l|c|}\hline \textbf{No}&\textbf{\: \: \qquad Bentuk Akar}&\textbf{\: \: \: \: Pangkat pecahan}\\\hline 1&\sqrt{a\times b}=\sqrt{a}\times \sqrt{b}&a^{n}=b\quad \Rightarrow \quad a=b^{^{^{\frac{1}{n}}}}=\sqrt[n]{b}\\\hline 2&\sqrt{\displaystyle \frac{a}{b}}=\displaystyle \frac{\sqrt{a}}{\sqrt{b}}&a> 0\quad \Rightarrow \quad \sqrt[n]{a}\geq 0\\\hline 3&a\sqrt{c}\pm b\sqrt{c}=\left ( a\pm b \right )\sqrt{c}&a< 0\begin{cases} n & \text{ genap } \Rightarrow \sqrt[n]{a}< 0\\ n & \text{ ganjil } \Rightarrow \sqrt[n]{a}\quad \textbf{bukan riil} \end{cases}\\\hline 4&\sqrt{\left ( a+b \right )\pm 2\sqrt{ab}}=\left (\sqrt{a} \pm \sqrt{b}\right )&a^{^{^{^{\frac{\begin{matrix} \\ m \end{matrix}}{\begin{matrix} n\\ \end{matrix}}}}}}=\sqrt[n]{a^{m}}\\\hline 5&\displaystyle \frac{a}{\sqrt{b}}=\frac{a}{\sqrt{b}}\times \frac{\sqrt{b}}{\sqrt{b}}=\frac{a}{b}\sqrt{b}&\\\cline{1-2} 6&\displaystyle \frac{c}{a\pm \sqrt{b}}=\frac{c}{a\pm \sqrt{b}}\times \frac{a\mp \sqrt{b}}{a\mp \sqrt{b}}=\frac{c\left ( a\mp \sqrt{b} \right )}{a^{2}- b}&\textrm{Perlu diketahui juga}\\\cline{1-2} 7&\begin{aligned}\displaystyle \frac{c}{\sqrt{a}\pm \sqrt{b}}&=\frac{c}{\sqrt{a}\pm \sqrt{b}}\times \frac{\sqrt{a}\mp \sqrt{b}}{\sqrt{a}\mp \sqrt{b}}\\ &=\frac{c\left ( \sqrt{a}\mp \sqrt{b} \right )}{a- b} \end{aligned}&a^{n}=b^{m}\Leftrightarrow a=\displaystyle b^{^{^{\frac{m}{n}}}}\Leftrightarrow a^{^{^{\frac{n}{m}}}}=b\\\cline{1-2} 8&\displaystyle \frac{a}{\sqrt[n]{b^{m}}}=\frac{a}{\sqrt[n]{b^{m}}}\times \frac{\sqrt[n]{b^{n-m}}}{\sqrt[n]{b^{n-m}}}=\frac{a}{b}\sqrt[n]{b^{n-m}}&\\\cline{1-2} 9&\begin{aligned}\displaystyle \frac{c}{\sqrt[3]{a}\pm \sqrt[3]{b}}=&\frac{c}{\sqrt[3]{a}\pm \sqrt[3]{b}}\\ &\times \frac{\sqrt[3]{a^{2}}\mp \sqrt[3]{ab}+\sqrt[3]{b^{2}}}{\sqrt[3]{a^{2}}\mp \sqrt[3]{ab}+\sqrt[3]{b^{2}}} \end{aligned}&\\\hline \end{array}.

D. Grafik Fungsi Eksponen

\LARGE{\fbox{\LARGE{\fbox{CONTOH SOAL}}}}.

\begin{array}{ll}\\ 1.&\textrm{Tentukanlah nilai dari bilangan-bilangan berikut ini}! \end{array}\\ \begin{array}{llllllll}\\ .\quad\quad &a.&27^{\frac{1}{3}}&k.&\left ( \displaystyle \frac{2^{3}.3^{-2}}{2^{-5}.3} \right )^{\displaystyle \frac{1}{2}}&u.&\displaystyle \frac{\left ( a^{2}.b^{-1} \right )^{\frac{1}{2}}\sqrt{a^{6}.b^{\frac{5}{3}}.c^{-2}}}{\left ( a^{3}.b^{-5}.c^{-3} \right )^{\frac{1}{3}}}\\ &b.&32^{^{\frac{2}{5}}}&l.&\displaystyle \frac{\sqrt{2}.\sqrt[3]{8}}{(2)^{\frac{1}{3}}.\sqrt[4]{16^{2}}}&v.&\displaystyle \frac{x^{2}.y^{7}}{x^{3}.y^{5}}\\ &c.&\left ( \displaystyle \frac{9}{16} \right )^{\displaystyle \frac{3}{2}}&m.&\left ( \displaystyle \frac{\sqrt{2}.2\sqrt{6}}{\sqrt{3}.\sqrt[3]{9}} \right )^{\displaystyle \frac{1}{2}}&w.&\left ( \displaystyle \frac{2x^{3}}{y^{2}}:\frac{4x^{6}}{4y^{5}} \right ).\displaystyle \frac{3x^{2}-2y}{3y}\\ &d.&\left ( \displaystyle \frac{1}{125} \right )^{-\frac{1}{3}}&n.&\displaystyle \frac{2.3^{-\frac{1}{2}}-2+3.2^{-\frac{1}{2}}}{2.3^{-\frac{1}{2}}-3.2^{-\frac{1}{2}}}&x.&\left (\sqrt[3]{x^{2}.yz^{3}} \right ).x^{-1}.y^{-2}\\ &e.&\left ( \displaystyle \frac{2}{3} \right )^{\displaystyle \frac{2}{3}}.\left ( \displaystyle \frac{3}{2} \right )^{-\displaystyle \frac{1}{3}}&o.&-3\sqrt{6}+4\sqrt{3}-2\sqrt{81}&y.&\displaystyle \frac{\sqrt{x}.\sqrt{x^{2}y^{3}}.\sqrt{xy^{2}}}{\sqrt[4]{x}.\sqrt[3]{y}}\\ &f.&\left ( \displaystyle \frac{1}{5^{3}} \right )^{-1}.\left ( \displaystyle \frac{1}{5^{2}} \right )^{2}&p.&\sqrt{250}-\sqrt{50}+15\sqrt{2}&z.&\displaystyle \frac{\left ( x^{2} \right )^{3}}{x^{4}}:\left ( \frac{x^{3}}{\left ( x^{3} \right )^{2}} \right )^{-2} \end{array}

\begin{array}{llllllll}\\ .\quad\quad&g.&\displaystyle \frac{\sqrt{3}.\sqrt{15}}{\sqrt{5}}\quad\qquad \: \: &q.&\sqrt[2]{75}-\sqrt[4]{27}+\sqrt[3]{128}\\ &h.&\displaystyle \frac{2\sqrt{3}.\sqrt{24}}{\sqrt{7}.3\sqrt{14}}&r.&\displaystyle \frac{5\sqrt{5}+2\sqrt{5}}{5-3\sqrt{5}}\\ &i&\displaystyle \frac{\sqrt{5}.\sqrt[2]{2^{3}}}{2.\sqrt[3]{3}}&s.&\sqrt{2\sqrt{2\sqrt{2\sqrt{2\sqrt{2\sqrt{\cdots }}}}}}\\ &j.&\displaystyle \frac{\sqrt{3}.\sqrt[3]{2}}{\left ( \frac{4}{9} \right )^{3}.\left ( \frac{2}{3} \right )^{-2}}&t.&\left ( 4^{\frac{1}{2}} \right )^{\frac{1}{2}}\left ( 2^{-2} \right )^{-2}.\sqrt[3]{0,125}.\left ( 0,25 \right ).\displaystyle \frac{1}{2}\sqrt{\frac{1}{2}} \end{array}.

. \quad\: \, \begin{aligned}&\textrm{Jawab}:\\ &\begin{array}{|l|l|l|l|}\hline \begin{aligned}a.\quad 27^{\frac{1}{3}}&=\left ( 3^{3} \right )^{\frac{1}{3}}\\ &=3\\ &\\ & \end{aligned}&\begin{aligned}b.\quad 32^{\frac{2}{5}}&=\left ( 2^{5} \right )^{\frac{2}{5}}\\ &=2^{2}=4\\ &\\ & \end{aligned}&\begin{aligned}c.\quad \left ( \displaystyle \frac{9}{16} \right )^{\frac{3}{2}}&=\left (\left ( \displaystyle \frac{3}{4} \right )^{2} \right )^{\frac{3}{2}}\\ &=\left ( \displaystyle \frac{3}{4} \right )^{3}=\frac{27}{64} \end{aligned}&\begin{aligned}d.\quad \left ( \displaystyle \frac{1}{125} \right )^{-\frac{1}{3}}&=\left ( 5^{-3} \right )^{-\frac{1}{3}}\\ &=5\\ & \end{aligned}\\\hline \begin{aligned}e.\quad &\left ( \frac{2}{3} \right )^{\frac{2}{3}}.\left ( \frac{3}{2} \right )^{-\frac{1}{3}}\\ &=\left ( \frac{2}{3} \right )^{\frac{2}{3}}.\left ( \frac{2}{3} \right )^{\frac{1}{3}}\\ &=\left ( \frac{2}{3} \right )^{\frac{2}{3}+\frac{1}{3}}\\ &=\frac{2}{3} \end{aligned}&\begin{aligned}h.\quad &\displaystyle \frac{2\sqrt{3}.\sqrt{24}}{\sqrt{7}.3\sqrt{14}}\\ &=\displaystyle \frac{2\sqrt{3}.\sqrt{4}.\sqrt{2}.\sqrt{3}}{\sqrt{7}.3.\sqrt{2}.\sqrt{7}}\\ &=\frac{4.\sqrt{2}.3}{7.\sqrt{2}.3}\\ &=\frac{4}{7} \end{aligned}&\begin{aligned}u.\quad &\displaystyle \frac{\left ( a^{2}b^{-1} \right )^{\frac{1}{2}}.\sqrt{a^{6}.b^{\frac{5}{3}}.c^{-2}}}{\left ( a^{3} \right )^{\frac{1}{3}}}\\ &=\displaystyle \frac{a.b^{-\frac{1}{2}}.a^{\frac{6}{2}}.b^{\frac{1}{2}.\frac{5}{3}}.c^{-\frac{2}{2}}}{a^{\frac{3}{3}}.b^{-\frac{5}{3}}.c^{-\frac{3}{3}}}\\ &=a^{3}.b^{-\frac{1}{2}+\frac{5}{6}+\frac{5}{3}}\\ &=a^{3}.b^{\frac{-3+5+10}{6}}=a^{3}.b^{2} \end{aligned}&\begin{aligned}&\textrm{Untuk Soal yang belum}\\ &\textrm{dibahas silahkan}\\ &\textrm{dikerjakan sendiri}\\ &\textrm{sebagai latihan} \end{aligned}\\\hline \end{array} \end{aligned}.

\begin{array}{ll}\\ 2.&\textrm{Tentukanlah himpunan penyelesaian (HP) dari}\\ &\textrm{a}.\quad 3^{2x-1}=1\\ &\textrm{b}.\quad 4^{x^{2}+3x-10}=1\\ &\textrm{c}.\quad 5^{3x^{2}+2x-1}=1\\ &\textrm{d}.\quad (9)^{2x+\frac{1}{2}}.\left ( \displaystyle \frac{1}{27} \right )^{3x^{2}+2}=1\\ &\textrm{e}.\quad \left ( \displaystyle \frac{1}{2} \right )^{2x-3}.(8)^{3x+1}=1\\ &\textrm{f}.\quad \displaystyle \frac{\sqrt[3]{\left ( 0,0008 \right )^{7-2x}}}{\left ( 0,2 \right )^{-4x+5}}=1\: \: ...(\textbf{SPMB 2005})\end{array}.

.\quad\: \, \begin{aligned}&\textrm{Jawab}:\\ &\begin{array}{|l|l|l|}\hline \begin{aligned}a.\quad 3^{2x-1}&=1\\ 3^{2x-1}&=3^{0}\\ 2x-1&=0\\ 2x&=1\\ x&=\displaystyle \frac{1}{2}\\ \textrm{HP}=&\left \{ \displaystyle \frac{1}{2} \right \}\\ &\\ &\\ & \end{aligned}&\begin{aligned}e.\quad \left ( \displaystyle \frac{1}{2} \right )^{2x-3}.(8)^{3x+1}&=1\\ (2^{3-2x}).(2^{3})^{3x+1}&=2^{0}\\ 2^{3-2x+9x+3}&=2^{0}\\ 7x+6&=0\\ 7x&=-6\\ x&=-\displaystyle \frac{6}{7}\\ \textrm{HP}=\left \{ -\displaystyle \frac{6}{7} \right \}&\\ & \end{aligned}&\begin{aligned}f.\quad \displaystyle \frac{\sqrt[3]{\left ( 0,0008 \right )^{7-2x}}}{\left ( 0,2 \right )^{-4x+5}}&=1\: \: ...(\textbf{SPMB 2005})\\ \displaystyle \frac{((0,2)^{3})^{\frac{7-2x}{3}}}{(0,2)^{-4x+5}}&=(0,2)^{0}\\ (0,2)^{7-2x-(-4x+5)}&=(0,2)^{0}\\ 7-2x+4x-5&=0\\ 2x+2&=0\\ 2&=-2\\ x&=-1\\ \textrm{HP}=\left \{ -1 \right \}& \end{aligned}\\\hline \end{array} \end{aligned}.

\begin{array}{ll}\\ 3.&\textrm{Tentukanlah himpunan penyelesaian (HP) dari}\\ &\textrm{a}.\quad 3^{2x-5}=3^{5}\\ &\textrm{b}.\quad 2^{2x+3}.\left (\displaystyle \frac{1}{8} \right )^{x-3}=\displaystyle \frac{1}{64}\\ &\textrm{c}.\quad (2)^{5x^{2}-3x}.\left ( \displaystyle \frac{1}{32} \right )^{5}=32^{5}\\ &\textrm{d}.\quad \left ( \displaystyle \frac{1}{9} \right )^{-x^{2}}.\left ( 3^{2} \right )^{3x-3}=\displaystyle 9^{3^{2}}\\ &\textrm{e}.\quad (4)^{-2x^{2}+3x}.\left ( \displaystyle \frac{1}{4} \right )^{3}=2^{-2}.\left ( \displaystyle \frac{1}{8} \right )^{-2}\\ &\textrm{f}.\quad \displaystyle \frac{27}{3^{2x-1}}=81^{-0,125}\: \: ...(\textbf{SPMB 2004})\end{array}.

.\quad\: \, \begin{aligned}&\textrm{Jawab}:\\ &\begin{array}{|l|l|}\hline \begin{aligned}a.\quad 3^{2x-5}&=3^{5}\\ a^{f(x)}&=a^{p}\\ f(x)&=p\\ 2x-5&=5\\ 2x&=10\\ x&=5\\ \textrm{HP}=&\left \{ 5 \right \}\\ &\\ &\\ &\\ &\\ &\end{aligned}&\begin{aligned}f.\quad \displaystyle \frac{27}{3^{2x-1}}&=81^{-0,125}\cdots (\textbf{SPMB 2004})\\ \displaystyle \frac{3^{3}}{3^{2x-1}}&=(3^{4})^{-\frac{1}{8}}\\ 3^{3-(2x-1)}&=3^{-\frac{4}{8}}\\ 3-(2x-1)&=-\frac{4}{8}\\ 4-2x&=-\frac{1}{2}\\ 8-4x&=-1\\ -4x&=-1-8\\ x&=\frac{9}{4}\\ \textrm{HP}=&\left \{ \frac{9}{4} \right \}\end{aligned}\\\hline \end{array} \end{aligned}.

\begin{array}{ll}\\ 4.&\textrm{Tentukanlah himpunan penyelesaian (HP) dari}\\ &\textrm{a}.\quad \sqrt{3^{2x+1}}=9^{x-2}\\ &\textrm{b}.\quad \left ( \displaystyle \frac{1}{3} \right )^{2x-3}.3^{x+5}=\left ( \displaystyle \frac{1}{27} \right )^{2x-10}\\ &\textrm{c}.\quad (125)^{(x^{2}-3x-4)}=(5)^{(x^{2}-2x-3)}\\ &\textrm{d}.\quad (2^{2})^{\displaystyle \sqrt{x^{3}+2x^{2}-3x-6}}=\left ( \displaystyle \frac{1}{2} \right )^{-\displaystyle \sqrt{4x^{2}+4x-8}}\\ &\textrm{e}.\quad (4)^{(x^{2}+2x-1)}.\left ( \displaystyle \frac{1}{8} \right )^{3x-4}=\left ( \displaystyle \frac{1}{2} \right )^{2x}\\ &\textrm{f}.\quad (\sqrt{2})^{4x^{2}-8x+12}.\left ( \sqrt[3]{8} \right )^{3x+5}=\left ( \displaystyle \frac{1}{4} \right )^{4x-3}.(2)^{6x+5}\\ &\textrm{g}.\quad 9^{x^{2}-3x+1}+9^{-3x+x^{2}}=20-10\left ( 3^{x^{2}-3x} \right )\: \: ...(\textbf{SPMB 2004})\end{array}.

.\quad\: \, \begin{aligned}&\textrm{Jawab}:\\ &\begin{array}{|l|l|l|}\hline \begin{aligned}a.\quad \sqrt{3^{2x+1}}&=9^{x-2}\\ a^{f(x)}&=a^{g(x)}\\ f(x)&=g(x)\\ \displaystyle (3)^{\frac{2x+1}{2}}&=(3^{2})^{x-2}\\ \displaystyle \frac{2x+1}{2}&=2(x-2)\\ 2x+1&=4x-8\\ -2x&=-9\\ x&=\displaystyle \frac{9}{2}\\ \textrm{HP}=&\left \{ \frac{9}{2} \right \} \end{aligned}&\begin{aligned}b.\quad \left ( \displaystyle \frac{1}{3} \right )^{2x-3}.3^{x+5}&=\left ( \displaystyle \frac{1}{27} \right )^{2x-10}\\ (3^{-1})^{(2x-3)}.3^{x+5}&=(3^{-3})^{(2x-10)}\\ 3^{(-2x+3)+(x+5)}&=3^{10-2x}\\ -2x+3+x+5&=10-2x\\ x&=10-8\\ x&=2\\ \textrm{HP}=\left \{ 2 \right \}&\\ &\\ &\\ & \end{aligned}&\begin{aligned}c.\quad (125)^{x^{2}-3x-4}&=(5)^{x^{2}-2x-3}\\ (5^{3})^{x^{2}-3x-4}&=(5)^{x^{2}-2x-3}\\ 3x^{2}-9x-12&=x^{2}-2x-3\\ 2x^{2}-7x-9&=0\\ (x+1)(2x-9)&=0\\ x=-1\: \: \textrm{V}\: \: x=\frac{9}{2}\\ \textrm{HP}=&\left \{ -1,\frac{9}{2} \right \}\\ &\\ &\\ & \end{aligned}\\\hline \multicolumn{3}{|l|}{\begin{aligned}g.\qquad\quad\quad\quad 9^{x^{2}-3x+1}+9^{-3x+x^{2}}&=20-10\left ( 3^{x^{2}-3x} \right )\\ \textrm{misalkan}\: \: 3^{x^{2}-3x}&=p\\ 9^{x^{2}-3x}.9+9^{x^{2}-3x}&=20-10\left ( 3^{x^{2}-3x} \right )\\ 9.\left ( 3^{2.(x^{2}-3x)} \right )+\left ( 3^{2.(x^{2}-3x)} \right )&=20-10\left ( 3^{x^{2}-3x} \right )\\ 10.\left ( 3^{x^{2}-3x} \right )^{2}&=20-10\left ( 3^{x^{2}-3x} \right )\\ 10p^{2}&=20-10p\\ p^{2}&=2-p\\ p^{2}+p-2&=0\\ (p+2)(p-1)&=0\\ p=-2\: (\textrm{tdk mungkin})\: \: \textrm{V}\: \: p=1&\\ \textrm{sehingga}\quad p=3^{x^{2}-3x}&=1\\ 3^{x^{2}-3x}&=3^{0}\\ x^{2}-3x&=0\\ x(x-3)&=0\\ x=1\: \: \textrm{V}\: \: x&=3\\ \textrm{HP}=&\left \{ 1,3 \right \} \end{aligned}}\\\hline \end{array} \end{aligned}.

\begin{array}{ll}\\ 5.&\textrm{Sederhanakanlah dengan merasionalkan penyebut-penyebutnya}\\ &a.\quad \displaystyle \frac{15}{\sqrt{5}}\qquad b.\quad \displaystyle \frac{9}{2\sqrt{3}}\, \qquad c.\quad \displaystyle \sqrt{\frac{7}{2}}\, \quad\qquad d.\quad \displaystyle \frac{3\sqrt{5}}{\sqrt{72}}\\ &e.\quad \displaystyle \frac{\sqrt{50}}{\sqrt{125}}\quad f.\quad \displaystyle \frac{3}{2-\sqrt{3}}\quad g.\quad \displaystyle \frac{6}{3+2\sqrt{3}}\quad h.\quad \displaystyle \frac{3-\sqrt{2}}{3+\sqrt{2}}\\\\ &\textrm{Jawab}:\\ &\begin{array}{|c|l|c|l|c|l|c|l|}\hline \multicolumn{2}{|c|}{\textrm{Pembahasan}}&\multicolumn{2}{|c|}{\textrm{Pembahasan}}&\multicolumn{2}{|c|}{\textrm{Pembahasan}}&\multicolumn{2}{|c|}{\textrm{Pembahasan}}\\\hline a.&\begin{aligned}\displaystyle \frac{15}{\sqrt{5}}&=\displaystyle \frac{15}{\sqrt{5}}\times \frac{\sqrt{5}}{\sqrt{5}}\\ &=\displaystyle \frac{15}{5}\sqrt{5}\\ &=3\sqrt{5} \end{aligned}&b.&\begin{aligned}\displaystyle \frac{9}{2\sqrt{3}}&=\displaystyle \frac{9}{2\sqrt{3}}\times \frac{\sqrt{3}}{\sqrt{3}}\\ &=\displaystyle \frac{9}{2\times 3}\sqrt{3}\\ &=\displaystyle \frac{3}{2}\sqrt{3} \end{aligned}&c.&\begin{aligned}\displaystyle \sqrt{\frac{7}{2}}&=\displaystyle \sqrt{\frac{7}{2}\times \frac{2}{2}}\\ &=\displaystyle \sqrt{\frac{14}{4}}\\ &=\displaystyle \frac{1}{2}\sqrt{14} \end{aligned}&d.&\begin{aligned}\displaystyle \frac{3\sqrt{5}}{\sqrt{72}}&=\displaystyle \frac{3\sqrt{5}}{\sqrt{72}}\times \frac{\sqrt{2}}{\sqrt{2}}\\ &=\displaystyle \frac{3\sqrt{5}.\sqrt{2}}{\sqrt{144}}\\ &=\displaystyle \frac{3\sqrt{10}}{12}\\ &=\displaystyle \frac{1}{4}\sqrt{10} \end{aligned} \\\hline \end{array} \end{array}..

.\quad\: \, \begin{aligned}&\begin{array}{|c|l|c|l|c|l|c|l|}\hline \multicolumn{2}{|c|}{\textrm{Pembahasan}}&\multicolumn{2}{|c|}{\textrm{Pembahasan}}&\multicolumn{2}{|c|}{\textrm{Pembahasan}}&\multicolumn{2}{|c|}{\textrm{Pembahasan}}\\\hline e.&\begin{aligned}\displaystyle \frac{50}{125}&=\displaystyle \frac{50}{125}\times \frac{5}{5}\\&=\displaystyle \frac{\sqrt{250}}{\sqrt{625}}\\ &=\displaystyle \frac{5\sqrt{10}}{25}\\ &=\displaystyle \frac{1}{5}\sqrt{10} \end{aligned}&f.&\begin{aligned}\displaystyle \frac{3}{2-\sqrt{3}}&=\displaystyle \frac{3}{2-\sqrt{3}}\times \frac{2+\sqrt{3}}{2+\sqrt{3}}\\ &=\displaystyle \frac{3\times \left ( 2+\sqrt{3} \right )}{2^{2}-\left (\sqrt{3} \right )^{2}}\\ &=\displaystyle \frac{6+3\sqrt{3}}{4-3}\\ &=6+3\sqrt{3} \end{aligned}&g.&\begin{aligned}\displaystyle \frac{6}{3+2\sqrt{3}}&=\displaystyle \frac{6}{3+2\sqrt{3}}\times \frac{3-2\sqrt{3}}{3-2\sqrt{3}}\\ &=\displaystyle \frac{6\times \left ( 3-2\sqrt{3} \right )}{3^{2}-\left (2\sqrt{3} \right )^{2}}\\ &=\displaystyle \frac{18-12\sqrt{3}}{9-12}\\ &=\displaystyle \frac{18-12\sqrt{3}}{-3}\\ &=-6+4\sqrt{3}\\ &=4\sqrt{3}-6 \end{aligned}&h.&\begin{aligned}\displaystyle &\frac{3-\sqrt{2}}{3+\sqrt{2}}\\ &=\displaystyle \frac{3-\sqrt{2}}{3+\sqrt{2}}\times \frac{3-\sqrt{2}}{3-\sqrt{2}}\\ &=\displaystyle \frac{3^{2}-2.3.\sqrt{2}+\left ( \sqrt{2} \right )^{2}}{3^{2}-\left ( \sqrt{2} \right )^{2}}\\ &=\displaystyle \frac{9+2-6\sqrt{2}}{9-2}\\ &=\displaystyle \frac{1}{7}\left ( 11-6\sqrt{2} \right ) \end{aligned}\\\hline \end{array} \end{aligned}.

\begin{array}{ll}\\ 6.&\textrm{Sederhanakanlah pecahan berikut dengan merasionalkan penyebutnya} \end{array}\\ \begin{array}{lllllllll}\\ .\quad\quad&a.&\displaystyle \frac{1}{\sqrt{5-2\sqrt{6}}}&b.&\displaystyle \frac{1}{\sqrt{15+\sqrt{224}}}&c.&\displaystyle \frac{1}{\sqrt{14-6\sqrt{5}}}-\frac{1}{\sqrt{14+6\sqrt{5}}}&d.&\displaystyle \frac{2}{\sqrt{7+2\sqrt{12}}}-\frac{1}{\sqrt{7+2\sqrt{12}}}\\ &&&&&&&\\ &e.&\displaystyle \frac{1}{\sqrt{2}+\sqrt{5}+\sqrt{7}}&f.&\displaystyle \frac{6}{\sqrt{11}-\sqrt{6}+\sqrt{5}}&g.&\displaystyle \frac{12}{3+\sqrt{7}-\sqrt{2}}&h.&\displaystyle \frac{5}{4-2\sqrt{3}+\sqrt{5}}\\ \end{array}.

.\quad\: \, \begin{aligned}&\begin{array}{|c|l|c|l|c|l|}\hline \multicolumn{2}{|c|}{\textrm{Pembahasan}}&\multicolumn{2}{|c|}{\textrm{Pembahasan}}&\multicolumn{2}{|c|}{\textrm{Pembahasan}}\\\hline a.&\begin{aligned}\displaystyle \frac{1}{\sqrt{5-2\sqrt{6}}}&=\displaystyle \frac{1}{5-2\sqrt{6}}\\ &=\displaystyle \frac{1}{\sqrt{3+2-2\sqrt{3.2}}}\\ &=\displaystyle \frac{1}{\sqrt{\left ( \sqrt{3}-\sqrt{2} \right )^{2}}}\\ &=\displaystyle \frac{1}{\sqrt{3}-\sqrt{2}}\\ &=\displaystyle \frac{1}{\sqrt{3}-\sqrt{2}}\times \frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}+\sqrt{2}}\\ &=\displaystyle \frac{\sqrt{3}+3\sqrt{2}}{3-2}\\ &=\sqrt{3}+\sqrt{2} \end{aligned}&b.&\begin{aligned}\displaystyle &\frac{1}{\sqrt{15+\sqrt{4.8.7}}}\\ &=\displaystyle \frac{1}{\sqrt{8+7+2\sqrt{8.7}}}\\ &=\displaystyle \frac{1}{\sqrt{\left ( \sqrt{8}+\sqrt{7} \right )^{2}}}\\ &=\displaystyle \frac{1}{\sqrt{8}+\sqrt{7}}\\ &=\displaystyle \frac{1}{\sqrt{8}+\sqrt{7}}\times \frac{\sqrt{8}-\sqrt{7}}{\sqrt{8}-\sqrt{7}}\\ &=\displaystyle \frac{\sqrt{8}-\sqrt{7}}{8-7}\\ &=\sqrt{8}-\sqrt{7} \end{aligned}&c.&\begin{aligned}\displaystyle &\frac{1}{\sqrt{14-6\sqrt{5}}}-\frac{1}{\sqrt{14+6\sqrt{5}}}\\ &=\displaystyle \frac{1}{\sqrt{14-2.3\sqrt{5}}}-\frac{1}{\sqrt{14+2.3\sqrt{5}}}\\ &=\displaystyle \frac{1}{\sqrt{9+5-2\sqrt{9.5}}}-\frac{1}{\sqrt{9+5+2\sqrt{9.5}}}\\ &=\displaystyle \frac{1}{\sqrt{9}-\sqrt{5}}-\displaystyle \frac{1}{\sqrt{9}+\sqrt{5}}\\ &=\displaystyle \left ( \frac{\sqrt{9}+\sqrt{5}}{9-5} \right )-\displaystyle \left ( \frac{\sqrt{9}-\sqrt{5}}{9-5} \right )\\ &=\displaystyle \frac{1}{4}\left ( 2\sqrt{5} \right )\\ &=\displaystyle \frac{1}{2}\sqrt{5} \end{aligned} \\\hline \end{array} \end{aligned}.

Sumber Referensi

  1. Kuntarti, Sulistiyono dan Kurnianingsih, S. 2005. Matematika untuk SMA dan MA Kelas XII Program Ilmu Alam. Jakarta: Gelora Aksara Pratama.
  2. Kuntarti, Sulistiyono dan Kurnianingsih, S. 2007. Matematika  SMA dan MA untuk Kelas XII Semester 2 Program IPA Standar Isi 2006. Jakarta: esis.
  3. Soetiyono, Sajaka, K. A., Suprijanto, S., Marwanta, Murniati, S., dan Herynugroho. 2007. Matematika Interaktif 3B Sekolah Menengah Atas Kelas XII Program Ilmu Pengetahuan Alam. Jakarta: Yudistira.
  4. Tim IGMP Matematika SMA. ….. . Tabloid Matematika Kurikulum 2006. Semarang: CV. Sarana Ilmu.
  5. Tung, K. Y. 2012. Pintar Matematika SMA Kelas XII IPA Untuk Olimpiade dan Pengayaan Pelajaran. Yogyakarta: ANDI.
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Fungsi Eksponen dan Fungsi Logaritma (Kelas X Peminatan)

Sebelumnya mari kita buka juga ke arsip lama berikut :

  • di sini 1 (materi fungsi, gunakan yang berkaitan dengan konsep fungsinya saja)
  • di sini 2 (materi eksponen/pangkat dan sifat-sifatnya serta contoh penggunaannya)
  • di sini 3 (operasi bentuk akar/pangkat bentuk rasional/pecahan)
  • di sini 4 (materi logaritma, sifat-sifatnya serta contoh penggunaannya)
  • di sini 5 (contoh soal-jawab ekponen/pangkat)
  • di sini 6 (materi dan contoh soal-jawab tentang ekponen dan fungsi eksponen serta logaritma)
  • di sini 7 (kumpulan contoh soal 1)
  • di sini 8 (kumpulan contoh soal 2)
  • di sini 9 (kumpulan contoh soal 3)
  • di sini 10 (kumpulan contoh soal 4)
  • di sini 11 (kumpulan contoh soal 5-logaritma)
  • di sini 12 (kumpulan contoh soal 6)
  • di sini 13 (kumpulan contoh soal 7-logaritma)
  • di sini 14 (contoh soal 1 disertai materi-KTSP 2006 kelas XII)
  • di sini 15 (contoh soal 2 disertai materi-KTSP 2006 kelas XII)
  • di sini 16 (contoh soal 3 disertai materi-KTSP 2006 kelas XII)
  • di sini 17 (kumpulan contoh 1 soal-K13 kelas X sebelum revisi)
  • di sini 18 (kumpulan contoh 2 soal-K13 kelas X sebelum revisi)
  • di sini 19 (contoh soal-jawab K13 kelas X revisi)

A. Bilangan Berpangkat Bulat Positif

Operasi bilangan berpangkat

\begin{array}{ll}\\ \multicolumn{2}{l}{\begin{aligned}\textrm{Yang}&\: \, \textrm{perlu diperhatikan untuk bilangan berpangkat/eksponen adalah}:\\ &\LARGE a^{m}=\displaystyle \underset{m\: \: \: \textbf{faktor}}{\underbrace{a\times a\times a\times a\times a\times a\times a\times a\times a\: \: ...\: \times a}}\\ &\bullet \: \: \: \: a^{m}\quad \textrm{disebut bilangan berpangkat}\\ &\bullet \: \: \: \: a\qquad \textrm{disebut bilangan pokok},\\ &\bullet \: \: \: \: n\qquad \textrm{disebut pangkat(eksponen)}\\ \textrm{Dan}\: \, &\textrm{sifat-sifat operasi bilangan berpangkat adalah sebagai berikut :} \end{aligned}}\\ 1.&a^{m}\times a^{n}=a^{(m+n)}\\ 2.&a^{m}\times b^{m}=(a\times b)^{m}\\ 3.&\displaystyle \frac{a^{m}}{a^{n}}=a^{(m-n)},\quad a\neq 0\\ 4.&\displaystyle \frac{a^{m}}{b^{m}}=\left ( \displaystyle \frac{a}{b} \right )^{m},\quad b\neq 0\\ 5.&\left ( a^{m} \right )^{n}=a^{(m\times n)}\\ 6.&a^{-m} =\displaystyle \frac{1}{a^{m}}, \: dan\: \: a^{m}=\displaystyle \frac{1}{a^{-m}},\quad a\neq 0\\ 7.&a^{1}=a\\ 8.&a^{0}=1,\quad \textrm{dengan}\: \: a\neq 0 \end{array}.

penjelasan untuk sifat di atas, di antaranya sebagai berikut :

\begin{aligned}\textbf{a}^{\textbf{m}}\: \times \: \textbf{a}^{\textbf{n}}&=\displaystyle \underset{...\: \: \: \textbf{faktor}}{\underbrace{\textbf{a}\times \textbf{a}\times \textbf{a}\times \textbf{a}\: \: ...\: \times \textbf{a}}}\: \: \: \times \: \: \: \displaystyle \underset{...\: \: \: \textbf{faktor}}{\underbrace{\: \: \qquad\begin{matrix} .\: .\: .\\ \end{matrix}\qquad\: \: }}\\ &=\displaystyle \underset{...\: \: \: \textbf{faktor}}{\underbrace{\: \: \qquad\begin{matrix} .\: .\: .\\ \end{matrix}\qquad\: \: }}\: \: \: \times \: \: \: \displaystyle \underset{...\: \: \: \textbf{faktor}}{\underbrace{\: \: \qquad\begin{matrix} .\: .\: .\\ \end{matrix}\qquad\: \: }}\\ &=\displaystyle \underset{...\: \: \: \textbf{faktor}}{\underbrace{\qquad\qquad\: \: \qquad\begin{matrix} .\: .\: .\\ \end{matrix}\qquad\: \: \qquad\qquad}}\\\\ &=\textbf{a}^{^{\textbf{...}}} \end{aligned}.

Sebagai misal

\begin{aligned}3^{2019}\: \times \: 3^{2020}&=\displaystyle \underset{...\: \: \: \textbf{faktor}}{\underbrace{3\times 3\times 3\times 3\: \: ...\: \times 3}}\: \: \: \times \: \: \: \displaystyle \underset{...\: \: \: \textbf{faktor}}{\underbrace{\: \: \qquad\begin{matrix} .\: .\: .\\ \end{matrix}\qquad\: \: }}\\ &=\displaystyle \underset{...\: \: \: \textbf{faktor}}{\underbrace{\: \: \qquad\begin{matrix} .\: .\: .\\ \end{matrix}\qquad\: \: }}\: \: \: \times \: \: \: \displaystyle \underset{...\: \: \: \textbf{faktor}}{\underbrace{\: \: \qquad\begin{matrix} .\: .\: .\\ \end{matrix}\qquad\: \: }}\\ &=\displaystyle \underset{...\: \: \: \textbf{faktor}}{\underbrace{\qquad\qquad\: \: \qquad\begin{matrix} .\: .\: .\\ \end{matrix}\qquad\: \: \qquad\qquad}}\\\\ &=...^{\: ^{...}} \end{aligned}.

dan untuk

\begin{aligned}\displaystyle \frac{a^{m}}{a^{n}}&=\displaystyle \frac{\overset{...\: \: \: \textbf{faktor}}{\overbrace{\overset{...\: \: \: \textbf{faktor}}{\overbrace{\qquad\: \begin{matrix} ...\\ \end{matrix}\: \qquad}}\: \: \times \: \: \overset{...\: \: \: \textbf{faktor}}{\overbrace{\qquad \: \begin{matrix} ...\\ \end{matrix}\: \qquad}}}}}{\displaystyle \underset{...\: \: \: \textbf{faktor}}{\underbrace{\qquad\qquad\: \: \qquad\begin{matrix} .\: .\: .\\ \end{matrix}\qquad\: \: \qquad\qquad}}}\\\\ &=\displaystyle \underset{...\: \: \: \textbf{faktor}}{\underbrace{\qquad\: \: \qquad\begin{matrix} .\: .\: .\\ \end{matrix}\qquad\: \: \qquad}}\\\\ &=\textbf{...}^{^{\textbf{...}}} \end{aligned}.

Sebagai contohnya adalah

\begin{aligned}\textbf{5}^{\textbf{2034}}\: : \: \textbf{5}^{\textbf{17}}&=\displaystyle \frac{\overset{2034\: \: \: \textbf{faktor}}{\overbrace{\overset{2017\: \: \: \textbf{faktor}}{\overbrace{\qquad\: \begin{matrix} ...\\ \end{matrix}\: \qquad}}\: \: \times \: \: \overset{17\: \: \: \textbf{faktor}}{\overbrace{\qquad \: \begin{matrix} ...\\ \end{matrix}\: \qquad}}}}}{\displaystyle \underset{17\: \: \: \textbf{faktor}}{\underbrace{\qquad\qquad\: \: \qquad\begin{matrix} .\: .\: .\\ \end{matrix}\qquad\: \: \qquad\qquad}}}\\\\ &=\displaystyle \underset{...\: \: \: \textbf{faktor}}{\underbrace{\qquad\: \: \qquad\begin{matrix} .\: .\: .\\ \end{matrix}\qquad\: \: \qquad}}\\\\ &=\textbf{...}^{^{\textbf{...}}} \end{aligned}.

Selanjutnya

\begin{aligned}\left ( \textbf{b}^{\textbf{m}} \right )^{\textbf{n}}&=\displaystyle \underset{...\: \: \: \textbf{faktor}}{\underbrace{\left ( \textbf{b}^{\textbf{m}} \right )\times \left ( \textbf{b}^{\textbf{m}} \right )\times \left ( \textbf{b}^{\textbf{m}} \right )\times \left ( \textbf{b}^{\textbf{m}} \right )\: \: ...\: \times \left ( \textbf{b}^{\textbf{m}} \right )}}\\ &=\displaystyle \underset{...\: \: \: \textbf{faktor}}{\underbrace{\: \: \qquad\begin{matrix} \displaystyle \underset{...\: \: \: \textbf{faktor}}{\underbrace{\: \: \qquad\begin{matrix} .\: .\: .\\ \end{matrix}\qquad\: \: }}\: \: \: \times \: \: \: \displaystyle \underset{...\: \: \: \textbf{faktor}}{\underbrace{\: \: \qquad\begin{matrix} .\: .\: .\\ \end{matrix}\qquad\: \: }}\: \: \: \times \: \: \: \displaystyle \underset{...\: \: \: \textbf{faktor}}{\underbrace{\: \: \qquad\begin{matrix} .\: .\: .\\ \end{matrix}\qquad\: \: }}\: \: \: \times ...\times \: \: \: \displaystyle \underset{...\: \: \: \textbf{faktor}}{\underbrace{\: \: \qquad\begin{matrix} .\: .\: .\\ \end{matrix}\qquad\: \: }}\: \: \: \\ \end{matrix}\qquad\: \: }}\\\\ &=\textbf{b}^{^{\textbf{...}}} \end{aligned}.

Untuk contohnya sebagai berikut

\begin{aligned}\left ( \textbf{4}^{\textbf{5}} \right )^{\textbf{2019}}&=\displaystyle \underset{...\: \: \: \textbf{faktor}}{\underbrace{\left ( \textbf{4}^{\textbf{5}} \right )\times \left ( \textbf{4}^{\textbf{5}} \right )\times \left ( \textbf{4}^{\textbf{5}} \right )\times \left ( \textbf{4}^{\textbf{5}} \right )\: \: ...\: \times \left ( \textbf{4}^{\textbf{5}} \right )}}\\\\ &=\displaystyle \underset{...\: \: \: \textbf{faktor}}{\underbrace{\: \: \qquad\begin{matrix} \displaystyle \underset{...\: \: \: \textbf{faktor}}{\underbrace{\: \: \qquad\begin{matrix} .\: .\: .\\ \end{matrix}\qquad\: \: }}\: \: \: \times \: \: \: \displaystyle \underset{...\: \: \: \textbf{faktor}}{\underbrace{\: \: \qquad\begin{matrix} .\: .\: .\\ \end{matrix}\qquad\: \: }}\: \: \: \times \: \: \: \displaystyle \underset{...\: \: \: \textbf{faktor}}{\underbrace{\: \: \qquad\begin{matrix} .\: .\: .\\ \end{matrix}\qquad\: \: }}\: \: \: \times ...\times \: \: \: \displaystyle \underset{...\: \: \: \textbf{faktor}}{\underbrace{\: \: \qquad\begin{matrix} .\: .\: .\\ \end{matrix}\qquad\: \: }}\: \: \: \\ \end{matrix}\qquad\: \: }}\\\\ &=\textbf{4}^{^{\textbf{...}}} \end{aligned}.

Dan

\begin{aligned}\left ( \textbf{p}\: \times \: \textbf{q} \right )^{\textbf{n}}&=\underset{...\: \: \: \textbf{faktor}}{\underbrace{\left ( \textbf{p}\: \times \: \textbf{q} \right )\times \left ( \textbf{p}\: \times \: \textbf{q} \right )\times \left ( \textbf{p}\: \times \: \textbf{q} \right )\times ...\times \left ( \textbf{p}\: \times \: \textbf{q} \right )}}\\\\ &=\underset{...\: \: \: \textbf{faktor}}{\underbrace{\qquad \begin{matrix} ...\\ \end{matrix}\qquad }}\: \: \: \times \: \: \: \underset{...\: \: \: \textbf{faktor}}{\underbrace{\qquad \begin{matrix} ...\\ \end{matrix}\qquad }} \\\\ &=... \end{aligned}.

Dan contohnya adalah

\begin{aligned}\left ( \textbf{9}\: \times \: \textbf{8} \right )^{\textbf{2019}}&=\underset{...\: \: \: \textbf{faktor}}{\underbrace{\left ( \textbf{9}\: \times \: \textbf{8} \right )\times \left ( \textbf{9}\: \times \: \textbf{8} \right )\times \left ( \textbf{9}\: \times \: \textbf{8} \right )\times ...\times \left ( \textbf{9}\: \times \: \textbf{8} \right )}}\\\\ &=\underset{...\: \: \: \textbf{faktor}}{\underbrace{\qquad \begin{matrix} ...\\ \end{matrix}\qquad }}\: \: \: \times \: \: \: \underset{...\: \: \: \textbf{faktor}}{\underbrace{\qquad \begin{matrix} ...\\ \end{matrix}\qquad }} \\\\ &=... \end{aligned}.

Berikutnya

\begin{aligned}\left ( \displaystyle \frac{\textbf{p}}{\textbf{q}} \right )^{\textbf{n}}&=\underset{...\: \: \: \textbf{faktor}}{\underbrace{\left ( \displaystyle \frac{\textbf{p}}{\textbf{q}} \right )\times \left ( \displaystyle \frac{\textbf{p}}{\textbf{q}} \right )\times \left ( \displaystyle \frac{\textbf{p}}{\textbf{q}} \right )\times ...\times \left ( \displaystyle \frac{\textbf{p}}{\textbf{q}} \right )}}\\\\ &=\displaystyle \frac{\overset{...\: \: \: \textbf{faktor}}{\overbrace{\qquad\qquad\begin{matrix} ...\\ \end{matrix}\qquad\qquad}}}{\underset{...\: \: \: \textbf{faktor}}{\underbrace{\qquad\qquad\begin{matrix} ...\\ \end{matrix}\qquad\qquad}}}\\\\ &=... \end{aligned}.

Contohnya

\begin{aligned}\left ( \displaystyle \frac{\textbf{3}}{\textbf{5}} \right )^{\textbf{2019}}&=\underset{...\: \: \: \textbf{faktor}}{\underbrace{\left ( \displaystyle \frac{\textbf{3}}{\textbf{5}} \right )\times \left ( \displaystyle \frac{\textbf{3}}{\textbf{5}} \right )\times \left ( \displaystyle \frac{\textbf{3}}{\textbf{5}} \right )\times ...\times \left ( \displaystyle \frac{\textbf{3}}{\textbf{5}} \right )}}\\\\ &=\displaystyle \frac{\overset{...\: \: \: \textbf{faktor}}{\overbrace{\qquad\qquad\begin{matrix} ...\\ \end{matrix}\qquad\qquad}}}{\underset{...\: \: \: \textbf{faktor}}{\underbrace{\qquad\qquad\begin{matrix} ...\\ \end{matrix}\qquad\qquad}}}\\\\ &=... \end{aligned}.

Untuk bentuk

\begin{aligned}\displaystyle \frac{\textbf{a}^{\textbf{m}}}{\textbf{a}^{\textbf{n}}}&=\displaystyle \frac{\overset{\textbf{m}\: \: \: \textbf{faktor}}{\overbrace{a\times a\times a\times a\times a\times a\times a\times a\times a\times ...\times a}}}{\underset{\textbf{n}\: \: \: \textbf{faktor}}{\underbrace{a\times a\times a\times a\times a\times a\times a\times a\times a\times a\times a\times a\times ...\times a}}},\quad \textnormal{dengan}\quad \textbf{m}< \textbf{n}\\ &=\displaystyle \frac{1}{\underset{\textbf{(n - m)}\: \: \: \textbf{faktor}}{\underbrace{a\times a\times a\times a\times \times a\times ...\times a}}}\\ &=\displaystyle \frac{1}{\textbf{a}^{\textbf{(n - m)}}},\qquad \textnormal{hal ini juga berarti}\\\\ &=\textbf{a}^{\textbf{(m - n)}}\\\\ &\qquad \textrm{yang perlu diingat bilangan dasar sama}\\\\ &\qquad \textrm{sehingga}\\\\ \textbf{a}^{\textbf{- p}}&=\displaystyle \frac{1}{\textbf{a}^{\textbf{p}}} \end{aligned}.

Contohnya

\begin{aligned}\displaystyle \frac{a^{6}b^{-4}c^{-1}d^{19}}{p^{7}q^{-3}r^{-1}s^{84}}&=\displaystyle \frac{a^{6}d^{19}b^{-4}c^{-1}}{p^{7}s^{84}q^{-3}r^{-1}}\\ &=\displaystyle \frac{a^{6}d^{19}}{p^{7}s^{84}}\times \displaystyle \frac{b^{-4}}{1}\times \displaystyle \frac{c^{-1}}{1}\times \displaystyle \frac{1}{q^{-3}}\times \frac{1}{r^{-1}}\\ &=\displaystyle \frac{a^{6}d^{19}}{p^{7}s^{84}}\times \frac{1}{b^{4}}\times \frac{1}{c^{1}}\times \frac{q^{3}}{1}\times \frac{r^{1}}{1}\\ &=\displaystyle \frac{a^{6}d^{19}q^{3}r^{1}}{b^{4}c^{1}p^{7}s^{84}} \\ &=\displaystyle \frac{a^{6}d^{19}q^{3}r}{b^{4}cp^{7}s^{84}}\\\\ &\qquad \textrm{perhatikanlah hasil akhir c berpangkat satu yang semula pangakt negatif 1},\\ &\qquad \textrm{sambil Anda perhatikan posisinya juga tentunya yaitu dari posisi pembilang berubah menjadi penyebut}.\\ &\qquad \textrm{Demikian pula r yang semual berpangkat negatif 1 berubah menjadi berpangkat positif satu}\\ &\qquad \textrm{dan posisinya berapada pada posisi pembilang}. \end{aligned}.

Perhatikan pula untuk bentuk

\begin{aligned}\displaystyle \frac{\textbf{b}^{m}}{\textbf{b}^{m}}&=1\\ \textrm{atau}&\\ 1&=\displaystyle \frac{\textbf{b}^{m}}{\textbf{b}^{m}}\\ 1&=\displaystyle \frac{\textbf{b}^{m}}{1}\times \frac{1}{\textbf{b}^{m}}\\ 1&=\textbf{b}^{m}\times \textbf{b}^{-m}\\ 1&=\textbf{b}^{(m - m)}\\ 1&=\textbf{b}^{0},\qquad\quad \textrm{dengan}\: \: b\neq 0 \end{aligned}.

B. Bilangan Eksponen Yang Berpangkat Berupa Fungsi

\begin{array}{|l|l|l|}\hline \textrm{No}&\textrm{Bentuk}&\textrm{Syarat}\\\hline 1.&a^{f(x)}=1&a\neq 0,\quad \textrm{maka}\: \: f(x)=0\\\hline 2.&a^{f(x)}=a^{p}&a>0,\: \: a\neq 1,\quad \textrm{maka}\: \: f(x)=p\\\hline 3.&a^{f(x)}=a^{g(x)}&a>0,\: \: a\neq 1,\quad \textrm{maka}\: \: f(x)=g(x)\\\hline 4.&a^{f(x)}=b^{f(x)}&a\neq 0,\: b\neq 0\: ,\quad \textrm{maka}\: \: f(x)=0\\\hline 5.&f(x)^{g(x)}=1&\begin{cases} f(x)=1 & \\ g(x)=0, & \textrm{jika}\: \: f(x)\neq 0 \\ f(x)=-1, & \textrm{jika}\: \: g(x)=\: \textrm{genap} \end{cases}\\\hline 6.&f(x)^{g(x)}=f(x)^{h(x)}&\begin{cases} (i).\quad g(x)=h(x)& \\ (ii).\quad f(x)=1& \\ (iii).\quad f(x)=0,&g(x)>0,\: \: h(x)>0 \\ (iv).\quad f(x)=-1,&g(x)\: \textrm{dan}\: h(x)\: \: \\ &\textrm{keduanya ganjil atau genap} \end{cases}\\\hline 7.&g(x)^{f(x)}=h(x)^{f(x)}&\begin{cases} (i).\quad g(x) =h(x)& \\ (ii).\quad f(x)=0, & g(x)\neq 0,\: h(x)\neq 0 \end{cases}\\\hline 8.&A\left ( a^{f(x)} \right )^{2}+B\left ( a^{f(x)} \right )+C=0&a>0,\: \: a\neq 1\\\hline \end{array}.

 

 

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