Bentuk Pangkat, Akar, dan Logaritma (KTSP)

Sebelumnya ada baiknya kita lihat lembaran yang sudah ada

  1. di sini
  2. di sini dan
  3. di sini serta
  4. contoh soalnya juga

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

\begin{array}{ll}\\ \fbox{1}.&\textrm{Tunjukkan}\: \textrm{bahwa}\: \textrm{untuk}\: \: m,\: n\: \textrm{bilangan}\: \textrm{bulat}\: \textrm{positif}\: \textrm{dan}\: \: p,\: q\: \in \mathbb{R}\: \textrm{berlaku}\\ &\textrm{a}.\quad p^{m}\times p^{n}=p^{m+n}\\ &\textrm{b}.\quad \left ( p.q \right )^{n}=p^{n}.q^{n}\end{array}.

Bukti:

\begin{aligned}1.\: \: \textrm{a}.\quad p^{m}\times p^{n}&=\overset{m\: \: \textrm{faktor}\: \: p}{\overbrace{p\times p\times p\times p\times ...\times p}}\times \overset{n\: \: \textrm{faktor}\: \: p}{\overbrace{p\times p\times p\times p\times ...\times p}}\\ &=\underset{\left (m+n \right )\: \: \textrm{faktor}\: \: p}{\underbrace{p\times p\times p\times p\times p\times p\times p\times p\times ...\times p}}\\ &=p^{m+n}\qquad \blacksquare \end{aligned}\\\\\\ \begin{aligned}1.\: \: \textrm{b}.\quad \left ( pq \right )^{n}&= \overset{n\: \: \textrm{faktor}\: \: pq}{\overbrace{pq\times pq\times pq\times pq\times pq\times ...\times pq}}\\ &=\underset{n\: \: \textrm{faktor}\: \: p}{\underbrace{p\times p\times p\times p\times p\times ...\times p}}\times \underset{n\: \: \textrm{faktor}\: \: q}{\underbrace{q\times q\times q\times q\times q\times ...\times q}}\\ &=p^{n}.q^{n}\qquad \blacksquare \end{aligned}.

\begin{array}{ll}\\ \fbox{2}.&\textrm{Tunjukkan}\: \textrm{bahwa}\: \textrm{untuk}\: \: m,\: n\: \textrm{bilangan}\: \textrm{bulat}\: \textrm{positif}\: \textrm{dan}\: \: p\in \mathbb{R}\left ( p\neq 0 \right ),\: \textrm{maka}\\\\ &\displaystyle \frac{p^{m}}{p^{n}}=\begin{cases} p^{m-n} &, \textrm{jika}\:\: m> n\quad ..........(i) \\ \displaystyle \frac{1}{p^{m-n}}&,\textrm{jika}\: \: m< n\quad ..........(ii) \\ 1 &, \textrm{jika}\: \: m=n\quad ..........(iii) \end{cases} \end{array}.

Bukti:

\begin{aligned}\displaystyle \frac{p^{m}}{p^{n}}&=\displaystyle \frac{\overset{m\: \: \textrm{faktor}\: p}{\overbrace{p\times p\times p\times p\times ...\times p}}}{\underset{n\: \: \textrm{faktor}\: p}{\underbrace{p\times p\times p\times p\times ...\times p}}}\\\\ &\textrm{Jika\: pada\: kasus}\: \left (\textrm{i} \right )\: \: m> n,\: \textrm{maka},\\\\ \displaystyle \frac{p^{m}}{p^{n}}&=\displaystyle \frac{\overset{n\: \: \textrm{faktor}\: p}{\overbrace{p\times p\times p\times p\times ...\times p}}\times \overset{\left (m-n \right )\: \: \textrm{faktor}\: p}{\overbrace{p\times p\times p\times p\times p\times ...\times p}}}{\underset{n\: \: \textrm{faktor}\: p}{\underbrace{p\times p\times p\times p\times ...\times p}}}\\ &=\displaystyle \frac{p^{n}\times p^{m-n}}{p^{n}}\\ &=p^{m-n}\qquad \blacksquare \end{aligned}\\\\ \textrm{Sedangkan\: untuk\: kasus}\: \left (\textrm{ii} \right )\: \textrm{dan}\: \left (\textrm{iii} \right )\: \textrm{silahkan\: dibuktikan\: sendiri} .

\begin{array}{ll}\\ \fbox{3}.&\textrm{Sederhanakanlah\: dengan\: menggunakan\: sifat-sifat\: bilangan\: berpangkat}\\ \end{array}\\ \begin{array}{lllllll}\\ .\: \: &\textrm{a}.& 3^{2}\times 3^{5}&\textrm{j}.&\left ( 2a^{3}b^{4} \right )^{8}\times \left ( 4ab^{3} \right )^{4}&\textrm{s}.&\left ( -h^{3} \right )^{4}\\ &\textrm{b}.& 4^{2}\times 4^{3}\times 4^{4}&\textrm{k}.&125^{2}\times 625\times 25^{2}&\textrm{t}.&\displaystyle \frac{\left (-k^{3} \right )^{2}}{k^{4}}\\ &\textrm{c}.& \displaystyle \frac{3^{7}}{3^{2}}&\textrm{l}.&16^{2}+\left ( 243-81^{3} \right )&\textrm{u}.&\displaystyle \frac{\left ( \displaystyle \frac{a^{2}b^{3}c^{5}}{ab} \right )^{3}\times \left ( \displaystyle \frac{a^{2}b^{3}}{b^{2}c^{4}} \right )}{\left ( \displaystyle \frac{a^{2}b}{c} \right )^{2}}\\ &\textrm{d}.& \displaystyle \frac{16a^{6}b^{8}}{2a^{3}b^{2}}&\textrm{m}.&\displaystyle \left ( \frac{3}{4} \right )^{4}\times \left ( \frac{2}{3} \right )^{3}&\textrm{v}.&\left ( \displaystyle \frac{1}{x-1} \right )\left ( \displaystyle \frac{1-x}{2-x} \right )^{2}\left ( \displaystyle \frac{x-2}{1-x} \right )^{3}\\ &\textrm{e}.& \left ( 3^{4} \right )^{6}&\textrm{n}.&\left ( 4^{6}+8^{3} \right )\times 2^{3}&&\\ &\textrm{f}.& \left ( 2x^{3}y^{2} \right )^{5}&\textrm{o}.&\left ( 0,25 \right )^{3}\times \left ( 0,125 \right )^{3}&&\\ &\textrm{g}.& \displaystyle \frac{\left (4a^{2}b^{4} \right )^{3}}{\left ( 2ab^{2} \right )^{2}}&\textrm{p}.&\displaystyle \frac{\left ( \displaystyle \frac{m^{3}n^{2}}{m} \right )^{6}}{\displaystyle \frac{m^{4}n^{3}}{mn^{2}}}&&\\ &\textrm{h}.&\displaystyle \frac{\left ( 2^{3} \right )^{4}\times \left ( 2^{2} \right )^{5}}{\left ( 2^{4} \right )^{2}}&\textrm{q}.&\left ( \displaystyle \frac{2ab^{4}}{ab^{2}} \right )^{3}\times \left ( \displaystyle \frac{a^{3}b^{4}}{2a^{2}b} \right )^{2}&&\\ &\textrm{i}.&\displaystyle \frac{27^{2}\times 3^{5}}{9^{2}}&\textrm{r}.&\left ( a^{2n}-7a^{n}+10 \right )\left ( a^{n}-1 \right )&& \end{array}.

Jawab:

Untuk jawaban  No  3  a  sampai  h  adalah sebagai berikut:

\begin{array}{|c|l|c|l|}\hline \multicolumn{2}{|c|}{\textbf{Pembahasan}\: \textbf{Soal}}&\multicolumn{2}{|c|}{\textbf{Pembahasan}\: \textbf{Soal}}\\\hline 3.a&3^{2}\times 3^{5}=3^{\left (2+5 \right )}=3^{5}&3.b&4^{2}\times 4^{3}\times 4^{4}=4^{\left (2+3+4 \right )}=4^{9}\\\hline 3.c&\displaystyle \frac{3^{7}}{3^{2}}=3^{\left (7-2 \right )}=3^{5}&3.d&\displaystyle \frac{16a^{6}b^{8}}{2a^{3}b^{2}}=8a^{\left (6-3 \right )}b^{\left (8-2 \right )}=8a^{3}b^{6}=2^{3}a^{3}b^{6}=\left ( 2ab^{2} \right )^{3}\\\hline 3.e&\left ( 3^{4} \right )^{6}=3^{\left (4\times 6 \right )}=3^{24}&3.f&\left ( 2x^{3}y^{2} \right )^{5}=2^{5}x^{15}y^{10}=32x^{15}y^{10}\\\hline 3.g&\begin{aligned}\displaystyle \frac{\left ( 4a^{2}b^{4} \right )^{3}}{\left ( 2ab^{2} \right )^{2}}&=\displaystyle \frac{4^{3}a^{6}b^{12}}{4a^{2}b^{4}}=4^{\left (3-1 \right )}a^{\left (6-2 \right )}b^{\left (12-4 \right )}\\ &=4^{2}a^{4}b^{8} \end{aligned}&3.h&\begin{aligned}\displaystyle \frac{\left ( 2^{3} \right )^{4}\times \left ( 2^{2} \right )^{5}}{\left ( 2^{4} \right )^{2}}&=\displaystyle \frac{2^{12}\times 2^{10}}{2^{8}}\\ &=2^{\left (12+10-8 \right )}\\ &=2^{14} \end{aligned}\\\hline \end{array}.

\begin{array}{|c|l|c|l|}\hline \multicolumn{2}{|c|}{\textbf{Pembahasan}\: \textbf{Soal}}&\multicolumn{2}{|c|}{\textbf{Pembahasan}\: \textbf{Soal}}\\\hline 3.i&\displaystyle \frac{27^{2}3^{5}}{9^{2}}=\frac{\left ( 3^{3} \right )^{2}.3^{5}}{\left ( 3^{2} \right )^{2}}=3^{3\times 2+5-2\times 2}=3^{6+5-4}=3^{7}&3.j&\begin{aligned} \left ( 2a^{3}b^{4} \right )^{8}\times \left ( 4ab^{3} \right )^{4}&=\left (2^{8}a^{24}b^{32} \right )\times \left ( 4^{4}a^{4}b^{12} \right )\\ &=2^{8}.\left ( 2^{2} \right )^{4}.a^{24+4}.b^{32+12}\\ &=2^{8+8}.a^{28}.b^{44}\\ &=2^{16}a^{28}b^{44}\end{aligned}\\\hline \end{array}.

\begin{array}{|c|l|c|l|}\hline \multicolumn{2}{|c|}{\textbf{Pembahasan}\: \textbf{Soal}}&\multicolumn{2}{|c|}{\textbf{Pembahasan}\: \textbf{Soal}}\\\hline 3.r&\begin{aligned}\left ( a^{2n}-7a^{n}+10 \right )\left ( a^{n}-1 \right )&=\left ( a^{n}-5 \right )\left ( a^{n}-2 \right )\left ( a^{n}-1 \right ) \end{aligned}&3.v&\begin{aligned}\left ( \displaystyle \frac{1}{x-1} \right )&\left ( \displaystyle \frac{1-x}{2-x} \right )^{2}\left ( \displaystyle \frac{x-2}{1-x} \right )^{3}\\ &=\left ( \displaystyle \frac{1}{x-1} \right ).\left ( \displaystyle \frac{x-1}{x-2} \right )^{2}.\displaystyle \frac{\left ( x-2 \right )^{3}}{-\left ( x-1 \right )^{3}}\\ &=-\displaystyle \frac{\left (x-1 \right )^{2}.\left ( x-2 \right )^{3}}{\left ( x-1 \right )^{4}.\left ( x-2 \right )^{2}}\\ &=-\left ( x-1 \right )^{2-4}.\left ( x-2 \right )^{3-2}\\ &=-\left ( x-1 \right )^{-2}.\left ( x-2 \right )\\ &=\displaystyle \frac{2-x}{\left ( x-1 \right )^{2}} \end{aligned}\\\hline \end{array}.

Untuk Jawaban yang lain silahkan dibuat sebagai latihan.

\begin{array}{ll}\\ \fbox{4}.&\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}.

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 4.a.&\begin{aligned}\displaystyle 2^{x}&=8\\ \displaystyle 2^{x}&=2^{3}\\ x&=3 \end{aligned}&4.b&\begin{aligned}\displaystyle 2^{x+1}&=32\\ \displaystyle 2^{x+1}&=2^{5}\\ x+1&=5\\ x&=4 \end{aligned}&4.c.&\begin{aligned}\displaystyle 2^{3x-4}&=32\\ \displaystyle 2^{3x-4}&=2^{5}\\ 3x-4&=5\\ 3x&=9\\ x&=3 \end{aligned}&4.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}&4.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}&4.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}.

\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 4.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}&4.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}&4.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}&4.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}&4.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}.

\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 4.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}&4.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}&4.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}.

\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 4.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}&4.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}&4.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}&4.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}.

Untuk soal-soal yang belum dibahas silahkan dibuat sebagai latihan.

\begin{array}{ll}\\ \fbox{5}.&\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}.

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}.

Untuk soal-soal yang belum dibahas silahkan dibuat sebagai latihan.

\begin{array}{lll}\\ \fbox{6}.&\textrm{Sederhanakanlah}\\ &&\\ &a.\quad \displaystyle \frac{3^{m+n}\times 3^{n+1}\times 3^{n-m}}{3^{2n+1}\times 3^{n+1}}&\qquad\qquad 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}} \end{array}.

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}.

Untuk soal-soal yang belum dibahas silahkan dibuat sebagai latihan.

\LARGE\fbox{\LARGE\fbox{LATIHAN SOAL}}.

  1. Tunjukkanlah sebagaimana perintah CONTOH SOAL no.2 kasus (ii) dan (iii).
  2. Selesaikanlah soal no.3 yang belum dibahas.
  3. Selesaikan juga soal no.4 yang belum dibahas.
  4. Selesaikan juga soal no.5 yang belum dibahas, serta
  5. Selesaikan juga soal no.6 yang belum dibahas.

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.

Tentang ahmadthohir1089

Nama saya Ahmad Thohir, asli orang Purwodadi lahir di Grobogan 02 Februari 1980. Pendidikan : Tingkat dasar lulus dari MI Nahdlatut Thullab di desa Manggarwetan,kecamatan Godong lulus tahun 1993. dan untuk tingkat menengah saya tempuh di MTs Nahdlatut Thullab Manggar Wetan lulus tahun 1996. Sedang untuk tingkat SMA saya menamatkannya di MA Futuhiyyah-2 Mranggen, Demak lulus tahun 1999. Setelah itu saya Kuliah di IKIP PGRI Semarang pada fakultas FPMIPA Pendidikan Matematika lulus tahun 2004. Pekerjaan : Sebagai guru (PNS DPK Kemenag) mapel matematika di MA Futuhiyah Jeketro, Gubug. Pengalaman mengajar : 1. GTT di MTs Miftahul Mubtadiin Tambakan Gubug tahun 2003 s/d 2005 2. GTT di SMK Negeri 3 Semarang 2005 s/d 2009 3. GT di MA Futuhiyah Jeketro Gubug sejak 1 September 2009
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