D. PERSAMAAN EKSPONEN
Berikut bentuk persamaan eksponen yang sering digunakan terangkum dalam tabel berikut beserta cara penyelesaiannya
$\begin{array}{|c|l|l|}\hline \textbf{No}&\textbf{Persamaan Eksponen}&\textbf{Penyelesaian}\\\hline 1&a^{f(x)}=1,\: \: a>0,a\neq 1&f(x)=0\\\hline 2&a^{f(x)}=a^{p},\: \: a>0,a\neq 1&f(x)=p\\\hline 3&a^{f(x)}=a^{g(x)},\: \: a>0,a\neq 1&f(x)=g(x)\\\hline 4&a^{f(x)}=b^{f(x)},\: \: a>0,a\neq 1&f(x)=0\\ &\qquad\quad \textrm{dan}\: \: b>0,\: b\neq 1&\\\hline 5&h(x)^{f(x)}=h(x)^{g(x)}&\begin{aligned}(1)\: &f(x)=g(x)\\ (2)\: &h(x)=1\\ (3)\: &h(x)=0\\ &\textrm{dengan syarat}\\ &f(x)> 0\: \: \textrm{dan}\\ &g(x)> 0\\ (4)\: &h(x)=-1\\ &\textrm{dengan syarat}\\ &f(x)\: \textrm{dan}\: g(x)\\ &\textrm{keduanya}\\ &\textrm{genap atau}\\ &\textrm{keduanya}\\ &\textrm{ganjil}\\ &\textrm{atau}\\ &\textrm{dapat juga}\\ &\textrm{ditunjukkan}\\ &(-1)^{f(x)}=(-1)^{g(x)} \end{aligned}\\\hline 6&g(x)^{f(x)}=h(x)^{f(x)}&\begin{aligned}(1)\: &g(x)=h(x)\\ (2)\: &f(x)=0\\ &\textrm{dengan syarat}\\ &g(x)\neq 0\: \: \textrm{dan}\\ &h(x)\neq 0\\ \end{aligned}\\\hline 7&f(x)^{g(x)}=1&\begin{aligned}(1)\: &f(x)=1\\ (2)\: &f(x)=-1\\ &\textrm{dengan syarat}\\ &g(x)\: \: \textrm{genap}\\ (3)\: &g(x)=0\\ &\textrm{dengan syarat}\\ &f(x)\neq 0 \end{aligned}\\\hline 8&A\left ( a^{f(x)} \right )^{2}+B\left ( a^{f(x)} \right )+C=0&\begin{aligned}&\textrm{ubah}\: \: a^{f(x)}=y\\ &\textrm{sehingga}\\ &Ay^{2}+By+C=0\\ &\textrm{selanjutnya}\\ &\textrm{substitusikan}\\ &\textrm{nilai}\: \: y\: \: \textrm{ke}\\ &\textrm{persamaan}\\ &a^{f(x)}=y \end{aligned}\\\hline \end{array}$.
$\LARGE{CONTOH SOAL}$.
$\begin{array}{ll}\\ 1.&\textrm{Tentukan himpunan penyelesaian dari}\\ &\textrm{a}.\quad 2^{2x-2021}=1\\ &\textrm{b}.\quad \left ( \displaystyle \frac{1}{2} \right )^{2x-2021}=1\\ &\textrm{c}.\quad \sqrt{2}^{2x-2021}=1\\\\ &\textbf{Jawab}:\\ &\begin{array}{|c|c|c|}\hline \textrm{a}&\textrm{b}&\textrm{c}\\\hline \begin{aligned} 2^{2x-2021}&=1\\ 2^{2x-2021}&=2^{0}\\ 2x-2021&=0\\ 2x&=2021\\ x&=\displaystyle \frac{2021}{2}\\ & \end{aligned}&\begin{aligned} \left ( \displaystyle \frac{1}{2} \right )^{2x-2021}&=1\\ \left ( \displaystyle \frac{1}{2} \right )^{2x-2021}&=\left ( \frac{1}{2} \right )^{0}\\ 2x-2021&=0\\ 2x&=2021\\ x&=\displaystyle \frac{2021}{2} \end{aligned}&\begin{aligned} \sqrt{2}^{2x-2021}&=1\\ \sqrt{2}^{2x-2021}&=\sqrt{2}^{0}\\ 2x-2021&=0\\ 2x&=2021\\ x&=\displaystyle \frac{2021}{2}\\ & \end{aligned}\\\hline \textbf{HP}=\left \{ \displaystyle \frac{2021}{2} \right \}&\textbf{HP}=\left \{ \displaystyle \frac{2021}{2} \right \}&\textbf{HP}=\left \{ \displaystyle \frac{2021}{2} \right \}\\\hline \end{array}\\ \end{array}$.
$\begin{array}{ll}\\ 2.&\textrm{Tentukan himpunan penyelesaian dari}\\ &\textrm{a}.\quad 2^{2x-2021}=128\\ &\textrm{b}.\quad \left ( \displaystyle \frac{1}{2} \right )^{2x-2021}=128\\ &\textrm{c}.\quad \sqrt{2}^{2x-2021}=128\\\\ &\textbf{Jawab}:\\ &\begin{array}{|c|c|c|}\hline \textrm{a}&\textrm{b}&\textrm{c}\\\hline \begin{aligned} 2^{2x-2021}&=128\\ 2^{2x-2021}&=2^{7}\\ 2x-2021&=7\\ 2x=7&+2021\\ x=\displaystyle \frac{2028}{2}&=1014\\ & \end{aligned}&\begin{aligned} \left ( \displaystyle \frac{1}{2} \right )^{2x-2021}&=128\\ \left ( \displaystyle \frac{1}{2} \right )^{2x-2021}&=\left ( \displaystyle \frac{1}{2} \right )^{-7}\\ 2x-2021&=-7\\ 2x=2021&-7\\ x=\displaystyle \frac{2014}{2}&=1007 \end{aligned}&\begin{aligned} \sqrt{2}^{2x-2021}&=128\\ \sqrt{2}^{2x-2021}&=\sqrt{2}^{256}\\ 2x-2021&=256\\ 2x=2021&+256\\ x&=\displaystyle \frac{2277}{2}\\ & \end{aligned}\\\hline \textbf{HP}=\left \{ 1014 \right \}&\textbf{HP}=\left \{ \displaystyle 1007 \right \}&\textbf{HP}=\left \{ \displaystyle \frac{2277}{2} \right \}\\\hline \end{array}\\ \end{array}$.
$\begin{array}{ll}\\ 3.&(\textbf{SPMB 04})\textrm{Nilai}\: \: x\: \: \textrm{yang memenuhi}\\ &\displaystyle \frac{27}{3^{2x-1}}=81^{-0,125} \: \: \textrm{adalah... .}\\\\ &\textbf{Jawab}:\\ &\begin{aligned}\displaystyle \frac{27}{3^{2x-1}}&=81^{-0,125}\\ 3^{3-(2x-1)}&=3^{4(\frac{1}{8})}\\ 3-2x+1&=-\displaystyle \frac{1}{2}\\ -2x+4&=-\displaystyle \frac{1}{2}\\ -x+2&=-\displaystyle \frac{1}{4}\\ -x&=-2-\displaystyle \frac{1}{4}\\ -x&=-2\displaystyle \frac{1}{4}\\ x&=2\displaystyle \frac{1}{4} \end{aligned} \end{array}$.
$\begin{array}{ll}\\ 4.&(\textbf{UMPTN 00})\\ &\textrm{Bentuk}\: \: \left (\sqrt[3]{\displaystyle \frac{1}{243}} \right )^{3x}=\left ( \displaystyle \frac{3}{3^{x-2}} \right )^{2}\sqrt[3]{\displaystyle \frac{1}{9}}\\ &\textrm{Jika}\: \: x_{0}\: \: \textrm{memenuhi persamaan, maka nilai}\\ &1-\displaystyle \frac{3}{4}x_{0}=\: ....\\\\ &\textbf{Jawab}:\\ &\begin{aligned}\left (\sqrt[3]{\displaystyle \frac{1}{243}} \right )^{3x}&=\left ( \displaystyle \frac{3}{3^{x-2}} \right )^{2}\sqrt[3]{\displaystyle \frac{1}{9}}\\ 3^{-5x}&=3^{2(1-(x-2))}.3^{-\frac{2}{3}}\\ -5x&=2(1-(x-2))+\left ( -\displaystyle \frac{2}{3} \right ),\: \: \textrm{dikali}\: \: 3\\ -15x&=6(3-x)+(-2)\\ -15x&=18-6x-2\\ 6x-15x&=16\\ -9x&=16\\ x&=\displaystyle \frac{16}{-9}\\ x_{0}&=-\displaystyle \frac{16}{9},\: \: \textrm{selanjutnya}\\ 1-\displaystyle \frac{3}{4}x_{0}&=1-\displaystyle \frac{3}{4}\times \left (-\frac{16}{9} \right )\\ &=1+\frac{4}{3}\\ &=1+1\displaystyle \frac{1}{3}\\ &=2\displaystyle \frac{1}{3} \end{aligned} \end{array}$.
$\begin{array}{l}\\ 5.&\textrm{Jumlah akar-akar persamaan}\\ & 5^{x+1}+5^{2-x}-30=0\: \: \textrm{adalah}\: ....\\\\ &\textbf{Jawab}:\\ &\begin{aligned}5^{x+1}+5^{2-x}-30&=0\\ \left (5^{x} \right ).5^{1}+\displaystyle \frac{5^{2}}{5^{x}}-30&=0\\ 5\left ( 5^{x} \right )^{2}+25-30\left ( 5^{x} \right )&=0\\ \textrm{Persamaan kuadrat}&\: \textrm{dalam}\: \: 5^{x},\: \textrm{maka}\\ 5(5^{x})^{2}-30(5^{x})+25&=0\begin{cases} a & =5 \\ b & =-30 \\ c & =25 \end{cases}\\ (5^{x_{1}}).\left ( 5^{x_{2}} \right )&=\displaystyle \frac{c}{a}\\ 5^{x_{1}+x_{2}}&=\displaystyle \frac{25}{5}=5\\ 5^{x_{1}+x_{2}}&=5^{1}\\ x_{1}+x_{2}&=1 \end{aligned} \end{array}$.
$\begin{array}{ll}\\ 6.&\textrm{Jumlah akar-akar persamaan}\\ &2023^{x^{2}-7x+7}=2024^{x^{2}-7x+7}\: \: \textrm{adalah}\: ....\\\\ &\textbf{Jawab}:\\ &\begin{aligned}2023^{x^{2}-7x+7}&=2024^{x^{2}-7x+7}\\ \textrm{Karena basis}&\: \textrm{tidak sama},\\ \textrm{maka harusl}&\textrm{ah pangkatnya}=0,\\ x^{2}-7x+7&=0\\ \textrm{dan jumlah}\: &\textrm{akar-akarnya adalah}:\\ x_{1}+x_{2}&=-\displaystyle \frac{b}{a}, \: \: \textrm{dari persamaan}\\ x^{2}-7x+7&=0\begin{cases} a &=1 \\ b &=-7 \\ c &=7 \end{cases}\\ \textrm{maka}\: \: x_{1}+x_{2}&=-\displaystyle \frac{b}{a}=-\frac{-7}{1}=7 \end{aligned} \end{array}$.
$\begin{array}{ll}\\ 7.&\textrm{Tentukan himpunan penyelesaian dari}\\ &(x-2)^{x^{2}-7x+6}=1\: \: \textrm{adalah}\: ....\\\\ &\textbf{Jawab}:\\ &\textrm{Ingat bentuk}\: \: \: f(x)^{g(x)}=1\begin{cases} f(x) & =x-2 \\ g(x) & =x^{2}-7x+5 \end{cases}\\ &\begin{array}{|l|l|l|}\hline f(x)=1&f(x)=-1&g(x)=0\\ &\textrm{Syarat}\: \: g(x)\: \: \textrm{genap}&\textrm{Syarat}\: \: f(x)\neq 0\\\hline \begin{aligned}x&-2=1\\ x&=3\\ & \end{aligned}&\begin{aligned}x&-2=-1\\ x&=2-1=1\\ & \end{aligned}&\begin{aligned}x^{2}&-7x+6=0\\ \Leftrightarrow &(x-1)(x-6)\\ \Leftrightarrow &\: x=1\: \: \textrm{atau}\: \: x=6 \end{aligned}\\\hline &\begin{aligned}\textrm{S}&\textrm{yaratnya}\: \: x\\ \textrm{u}&\textrm{ntuk}\: \: x=1\\ g&(1)=1^{2}-7+6\\ &=0\: \: (\textrm{memenuhi}) \end{aligned}&\begin{aligned}f(1)&=1-2=-1\neq 0\\ f(6)&=6-2=4\neq 0\\ &\\ & \end{aligned}\\ &\textbf{Catatan}:\: 0&\\ &\textrm{paritasnya genap}&\\\hline \end{array}\\ &\textbf{HP}=\left \{ 1,3,6 \right \} \end{array}$.
$\begin{array}{ll}\\ 8.&\textrm{Tentukan himpunan penyelesaian dari}\\ &(x^{2}-9x+19)^{2x+3}=(x^{2}-9x+19)^{x-1}\: \: \textrm{adalah}\: ....\\\\ &\textbf{Jawab}:\\ &\textrm{Ingat bentuk}\: \: \: h(x)^{f(x)}=h(x)^{g(x)}\begin{cases} h(x) & =x^{2}-9x+19 \\ f(x) & =2x+3\\ g(x)&=x-1 \end{cases}\\ &\begin{array}{|l|l|}\hline \begin{aligned}&\textrm{Syarat-syaratnya}\\ &\bullet \: \: f(x)=g(x)\\ &\Leftrightarrow 2x+3=x-1\\ &\Leftrightarrow x=-4\\ &\bullet \: \: h(x)=1\\ &\Leftrightarrow x^{2}-9x+19=1\\ &\Leftrightarrow x^{2}-9x+18=0\\ &\Leftrightarrow (x-3)(x-6)=0\\ &\Leftrightarrow x=3\: \: \textrm{atau}\: \: x=6\\ &\bullet \: \: h(x)=0\\ &\Leftrightarrow x^{2}-9x+19=0\\ &\Leftrightarrow x_{1,2}=\displaystyle \frac{9\pm \sqrt{5}}{2}\\ &\quad \textrm{gunakan rumus ABC}\\ &\textrm{Setelah diuji keduanya}\\ &\textrm{positif, maka}\\ &x=\displaystyle \frac{9\pm \sqrt{5}}{2}\: \: \textrm{merupakan}\\ &\textbf{penyelesaian} \end{aligned} &\begin{aligned}&\textrm{lanjutannya}\\ &\bullet \: \: h(x)=-1\\ &\Leftrightarrow x^{2}-9x+19=-1\\ &\Leftrightarrow x^{2}-9x+20=0\\ &\Leftrightarrow (x-4)(x-5)=0\\ &\Leftrightarrow x=4\: \: \textrm{atau}\: \: x=5\\ &\textrm{Uji nilanya}\\ &\textrm{untuk}\: \: x=4\\ &\blacklozenge \: \: f(4)=2(4)+3\: \: \textrm{ganjil}\\ &\blacklozenge \: \: g(4)=4-1\: \: \textrm{ganjil}\\ &\textrm{karena}\: f(4),g(4)\: \textrm{keduanya }\\ &\textrm{ganjil, maka}\: \: x=4\\ &\textrm{adalah}\: \textbf{penyelesaian} \\ &\textrm{untuk}\: \: x=5\\ &\blacklozenge \: \: f(5)=2(5)+3\: \: \textrm{ganjil}\\ &\blacklozenge \: \: g(5)=5-1\: \: \textrm{genapl}\\ &\textrm{karena}\: f(4)\neq g(4),\: \textrm{maka}\: \: x=5\\ &\textrm{adalah}\: \textbf{bukan penyelesaian}\\ &\\ \end{aligned}\\\hline \end{array}\\ &\textbf{HP}=\left \{ -4,3,4,6,\displaystyle \frac{9-\sqrt{5}}{2},\frac{9+\sqrt{5}}{2} \right \} \end{array}$.
DAFTAR PUSTAKA
- Kurnia, N, dkk. 2016. Jelajah Matematika I SMA Kelas X Peminatan MIPA. Jakarta: YUDHISTIRA.
