Lanjutan Contoh Soal Eksponens dan Logaritma(KTSP Kelas XII) Bagian 2

\begin{array}{ll}\\ \fbox{5}.&\textrm{Tentukanlah himpunan penyelesaian (HP) dari}\\ &\textrm{a}.\quad 3^{2x-5}=15^{2x-5}\\ &\textrm{b}.\quad 3^{2x-5}=15^{x+5}\\ &\textrm{c}.\quad 3^{2x-8}=4^{2x-8}\\ &\textrm{d}.\quad 5^{2x-8}=6^{x+3}\\ &\textrm{e}.\quad 7^{x^{2}-2x-8}=8^{x^{2}-2x-8}\\ &\textrm{f}.\quad 9^{x^{2}-2x-8}=10^{x^{2}-2x-8} \end{array}.

Jawab:

\begin{array}{|l|l|}\hline \begin{aligned}a.\quad 3^{2x-5}&=15^{2x-5}\\ a^{f(x)}&=b^{f(x)}\\ f(x)&=0\\ 2x-5&=0\\ x&=\displaystyle \frac{5}{2}\\ \textrm{HP}=&\left \{ \displaystyle \frac{5}{2} \right \}\\ &\\ &\\ &\\ & \end{aligned}&\begin{aligned}b.\quad\quad\quad\quad\quad\quad 3^{2x-5}&=15^{x+5}\\ a^{f(x)}&=b^{g(x)}\qquad \textrm{di\textit{log}kan masing-masing ruas}\\ \log 3^{(2x-5)}&=\log 15^{(x+5)}\\ (2x-5)\log 3&=(x+5)\log \left (3\times 5 \right )\\ (2x-5)\log 3&=(x+5)\left ( \log 3+\log 5 \right )\\ (2x-5)(0,4771)&=(x+5)(0,4771+0,699)\\ 0,9542x-2,3855&=1,1761x+5,8805\\ 0,9542x-1,1761x&=5,8805+2,3855\\ -0,2219x&=8,266\\ x&=\displaystyle \frac{8,266}{-0,2219}=-37,251\\ \textrm{HP}=&\left \{ -37,251 \right \} \end{aligned}\\\hline \end{array}.

\begin{array}{ll}\\ \fbox{6}.&\textrm{Jika terdapat hubungan berikut}\\ &\textrm{a}.\quad 2^{p}=3^{q}=6^{r},\: \: \textrm{tunjukkan bahwa akan didapatkan}\: \: pr+qr-pq=0\\ &\textrm{b}.\quad 2^{x}=3^{2y}=6^{z},\: \: \textrm{tunjukkan bahwa akan didapatkan}\: \: 2xy-2yz-xz=0\\ &\textrm{c}.\quad 3^{15a}=5^{5b}=15^{3c},\: \: \textrm{tunjukkan bahwa akan didapatkan}\: \: 5ab-bc-3ac=0\\ \end{array}.

Bukti:

Yang ditunjukkan hanya no.6 c saja

\begin{aligned}3^{15a}&=5^{5b}=15^{3c}\begin{cases} 3=5^{\frac{5b}{15a}} & \\ 3^{\frac{15a}{5b}}=b &\left ( a^{b}=c^{d}\rightarrow a=c^{\frac{d}{b}}\: \: \textrm{atau}\: \: a^{\frac{b}{d}}=c \right ) \end{cases}\\ 3^{15a}&=15^{3c}\\ 3^{15a}&=(3\times 5)^{3c}\\ 3^{15a}&=(3\times 3^{\frac{15a}{5b}})^{3c}\\ 3^{15a}&=3^{3c+\frac{9c}{b}}\\ a^{f(x)}&=a^{g(x)}\\ f(x)&=g(x)\\ 15a&=3c+\frac{9ac}{b}\\ 15ab&=3bc+9ac\\ 5ab&=bc+3ac\\ 5ab-bc-3ac&=0\quad \blacksquare \end{aligned}.

\begin{array}{ll}\\ \fbox{7}.&\textrm{Tentukanlah himpunan penyelesaian (HP) dari}\\ &\textrm{a}.\quad (x-7)^{2x-9}=(x-7)^{x+5}\\ &\textrm{b}.\quad (3x-5)^{x-4}=(3x-5)^{x+1}\\ &\textrm{c}.\quad (x-6)^{x^{2}-5}=(x-6)^{2x+3}\\ &\textrm{d}.\quad (5x-6)^{x^{2}-2}=(5x-6)^{4x+3}\\ &\textrm{e}.\quad (x-2)^{3x^{2}+5x+1}=(x-2)^{x^{2}-2x-2}\\ &\textrm{f}.\quad (x^{2}+3x-16)^{x+3}=(x^{2}+3x-16)^{2x+1}\\ &\textrm{g}.\quad (x^{2}-9x+19)^{2x+3}=(x^{2}-9x+19)^{x-1} \end{array}.

Jawab:

Yang dibahas hanya soal No. 7 c saja, yaitu:

\begin{array}{|l|l|l|l|} \multicolumn{4}{c}{c.\quad (x-6)^{x^{2}-5}=(x-6)^{2x+3}}\\ \multicolumn{4}{c}{\overset{\Downarrow }{f(x)^{g(x)}=f(x)^{h(x)}}}\\\hline \multicolumn{2}{|l}{.}&\multicolumn{2}{|l|}{.}\\\cline{2-3} \begin{aligned}x^{2}-5&=2x+3\\ x^{2}-2x-8&=0\\ (x-4)(x+2)&=0\\ x=4\: \textrm{V}\: x=-2 \end{aligned}&g(x)=h(x)&f(x)=0&\begin{aligned}x-6&=0\\ x&=6\\ g(6)&=(6)^{2}-5=31>0\\ h(6)&=2(6)+3=15>0\\ \textrm{maka}\: &x=6\: \: \textrm{memenuhi} \end{aligned}\\\hline \begin{aligned}x-6&=1\\ x&=7 \end{aligned}&f(x)=1&f(x)=-1&\begin{aligned}x-6&=-1\\ x&=5\\ g(5)&=5^{2}-5=20(genap)\\ h(5)&=2(5)+3=13(ganjil)\\ \textrm{maka} \: &x=5\: \: \textrm{tidak memenuhi} \end{aligned}\\\cline{2-3} \multicolumn{2}{|l}{.}&\multicolumn{2}{|l|}{.}\\\hline \end{array}\\\\ \: \textrm{Jadi},\\ \qquad \textrm{HP}=\left \{-2,4,6,7 \right \}.

\begin{array}{ll}\\ \fbox{8}.&\textrm{Tentukan himpunan penyelesaian (HP) dari}\\ &\begin{array}{llll}\\ \textrm{a}.\quad (6x-5)^{x+7}=(3x+1)^{x+7}&\textrm{j}.\quad (x^{2}-10)^{x^{2}-9x+14}=1\\ \textrm{b}.\quad (x+1)^{x^{2}-3x}=(2x-3)^{x^{2}-3x}&\textrm{k}.\quad (3x^{2}-13x+3)^{9x^{2}-5}=1\\ \textrm{c}.\quad (1-2x)^{x^{2}-4x+3}=(1+2x)^{x^{2}-4x+3}&\textrm{l}.\quad 4^{x}-10.2^{x}=-16\\ \textrm{d}.\quad (x^{2}+3x-2)^{x^{3}-x}=(2x^{2}-4x+8)^{x^{3}-x}&\textrm{m}.\quad 3^{2x+1}-23=\displaystyle \frac{36}{3^{2x}}\\ \textrm{e}.\quad (3x^{2}-5x+1)^{x^{2}-4x+3}=(x^{2}+4x-3)^{x^{2}-4x+3}&\textrm{n}.\quad 4^{x+1}+11.2^{x}=3\\ \textrm{f}.\quad (x-5)^{x+2}=1&\textrm{o}.\quad 2^{2x+2}-18.2^{x}+8=0\\ \textrm{g}.\quad (2x+6)^{x^{2}-4}=1&\textrm{p}.\quad 3^{x+1}+9^{x+1}=12\\ \textrm{h}.\quad (x^{2}-x-1)^{x+2}=1\: ....(\textbf{OSK 2005})&\textrm{q}.\quad 5^{x+1}-5^{1-x}+24=0 \\ \textrm{i}.\quad (x^{2}-3x-11)^{x+1}=1&\textrm{r}.\quad 3.7^{2-x}-7^{3-x}=20\\\end{array} \end{array}.

Jawab:

Yang dibahas hanya No. 8 a, 8f, dan 8p, yaitu:

\begin{array}{|l|l|}\hline \begin{aligned}(6x-5)^{x+7}&=(3x+1)^{x+7}\\ g(x)^{f(x)}&=h(x)^{f(x)}\\ \textcircled{1}\qquad g(x)&=h(x)\\ 6x-5&=3x+1\\ 3x&=6\\ x&=2\\ \textcircled{2}\qquad f(x)&=0\\ x+7&=0\\ x&=-7\\ &(i)\quad g(0)=6.0-5=-5\neq 0\\ &(ii)\: \: \: h(x)=3.0+1=1\neq 0\\ \textrm{HP}=&\left \{ -7,2 \right \} \end{aligned}&\begin{aligned}(x-5)^{x+2}&=1\\ f(x)^{g(x)}&=1\\ \textcircled{1}\qquad f(x)&=x-5=1\\ &\: \quad x=6\\ \textcircled{2}\qquad f(x)&=x-5=-1\\ &\: \quad x=4\: \Rightarrow g(4)=4+2=6(genap)\\ \textcircled{3}\qquad g(x)&=x+2=0\\ &\: \quad x=-2\: \Rightarrow f(-2)=-2-5=-7\neq 0\\ \textrm{HP}=&\left \{ -2,4,6 \right \}\\ &\\ &\\ & \end{aligned}\\\hline \multicolumn{2}{|c|}{\begin{aligned}3^{x+1}+9^{x+1}&=12\\ 3^{x}.3+(3^{2})^{x+1}&=12\\ 3.3^{x}+(3^{x})^{2}.3^{2}-12&=0,\qquad \textrm{misalkan}\: \: 3^{x}=p\\ 3p+9p^{2}-12&=0\\ 3p^{2}+p-4&=0\\ (p-1)(3p+4)&=0\\ \left ( 3^{x}-1 \right )\left ( 3.3^{x}+4 \right )&=0\\ 3^{x}=1\: \: atau\: \: 3^{x}&=-\displaystyle \frac{4}{3}\: (\textrm{tidak mungkin})\\ 3^{x}&=1\\ 3^{x}&=3^{0}\\ x&=0\\ \textrm{HP}=&\left \{ 0 \right \} \end{aligned}}\\\hline \end{array}.

\begin{array}{ll}\\ \fbox{9}.&\textrm{Tentukanlah penyelesaian dari tiap pertidaksamaan eksponen berikut}\\ &\textrm{a}.\quad 4^{x+1}>32\\ &\textrm{b}.\quad 8^{x+1}>4^{x}\\ &\textrm{c}.\quad 36^{x-3}<\left (\sqrt{6} \right )^{2x+1}\\ &\textrm{d}.\quad \left ( \displaystyle \frac{1}{3} \right )^{5x-2}<\left ( \displaystyle \frac{1}{81} \right )^{2x+1}\\ &\textrm{e}.\quad \left ( \displaystyle \frac{1}{8} \right )^{2x-3}>\left ( \displaystyle \frac{1}{32} \right )^{x+2}\\ &\textrm{f}.\quad \left ( \displaystyle \frac{1}{8\sqrt{2}} \right )^{x}\leq \left ( \displaystyle \frac{1}{16} \right )^{x-2}\\ &\textrm{g}.\quad (0,5)^{x^{2}-8x+7}>1\\ &\textrm{h}.\quad \left ( \displaystyle \frac{27}{125} \right )^{x-5}<\displaystyle \frac{625}{81}\\ &\textrm{i}.\quad \displaystyle \sqrt[3]{32\sqrt{8^{x+1}}}>\displaystyle \frac{1}{4^{3-x}}\\ &\textrm{j}.\quad 3^{x-2}<1<5^{2x+8} \end{array}.

Jawab:

Yang dibahas hanya soal No. 9a dan 9d, yaitu:

\begin{array}{|l|l|}\hline \begin{aligned}4^{x+1}&>32\\ (2^{2})^{x+1}&>2^{5}\\ 2(x+1)&>5\\ 2x+2&>5\\ 2x&>5-2\\ x&>\displaystyle \frac{3}{2}\\ &\\ & \end{aligned}&\begin{aligned}\left ( \displaystyle \frac{1}{3} \right )^{5x-2}&<\left ( \displaystyle \frac{1}{81} \right )^{2x+1}\\ \left ( \displaystyle \frac{1}{3} \right )^{5x-2}&<\left ( \displaystyle \left (\frac{1}{3} \right )^{4} \right )^{2x+1}\\ 5x-2&>4(2x+1)\\ 5x-8x&>4+2\\ -3x&>6\\ x&<- 2 \end{aligned}\\\hline \end{array}.

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

Kerjakanlah soal-soal  pada No. 1 sampai dengan No.9 yang belum dibahas atau belum ditunjukkan buktinya.

Sumber Referensi

  1. Soetiyono, Kamta Agus Sajaka, Sigit suprijanto, Marwanta, Suwarsini Murniati, dan Herynugroho. 2007. Matematika Interaktif 3B Sekolah Menengah Atas Kelas XII Program Ilmu Pengetahuan Alam. Jakarta: Yudistira.
  2. Tim IGMP Matematika SMA. ….. . Tabloid Matematika Kurikulum 2006. Semarang: CV. Sarana Ilmu.
  3. Tung, Khoe Yao. 2012. Pintar Matematika SMA Kelas XII IPA Untuk Olimpiade dan Pengayaan Pelajaran. Yogyakarta: ANDI.

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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