Lanjutan Materi : Persamaan Eksponen

A. Persamaan Eksponen

Berikut bentuk persamaan eksponen yang sering digunakan terangkum dalam tabel berikut beserta cara penyelesaiannya

$\begin{array}{|cll|}\hline \mathbf{\text{No}}& \mathbf{\text{Persamaan Eksponen}}& \mathbf{\text{Penyelesaian}}\\ 1& a^{f(x)}=1,a>0,a\ne 1& f(x)=0\\ 2& a^{f(x)}=a^p,a>0,a\ne 1& f(x)=p\\ 3& a^{f(x)}=a^{g(x)},a>0,a\ne 1& f(x)=g(x)\\ 4& a^{f(x)}=b^{f(x)},a>0,a\ne 1& f(x)=0\\ & \text{dan}b>0,b\ne 1& \\ 5& h(x{)}^{f(x)}=h(x{)}^{g(x)}& \begin{array}{rl}(1)& f(x)=g(x)\\ (2)& h(x)=1\\ (3)& h(x)=0\\ & \text{dengan syarat}\\ & f(x)>0\text{dan}\\ & g(x)>0\\ (4)& h(x)=-1\\ & \text{dengan syarat}\\ & f(x)\text{dan}g(x)\\ & \text{keduanya}\\ & \text{genap atau}\\ & \text{keduanya}\\ & \text{ganjil}\\ & \text{atau}\\ & \text{dapat juga}\\ & \text{ditunjukkan}\\ & (-1{)}^{f(x)}=(-1{)}^{g(x)}\end{array}\\ 6& g(x{)}^{f(x)}=h(x{)}^{f(x)}& \begin{array}{rl}(1)& g(x)=h(x)\\ (2)& f(x)=0\\ & \text{dengan syarat}\\ & g(x)\ne 0\text{dan}\\ & h(x)\ne 0\end{array}\\ 7& f(x{)}^{g(x)}=1& \begin{array}{rl}(1)& f(x)=1\\ (2)& f(x)=-1\\ & \text{dengan syarat}\\ & g(x)\text{genap}\\ (3)& g(x)=0\\ & \text{dengan syarat}\\ & f(x)\ne 0\end{array}\\ 8& A{\left(a^{f(x)}\right)}^2+B\left(a^{f(x)}\right)+C=0& \begin{array}{rl}& \text{ubah}{a}^{f(x)}=y\\ & \text{sehingga}\\ & A{y}^2+By+C=0\\ & \text{selanjutnya}\\ & \text{substitusikan}\\ & \text{nilai}y\text{ke}\\ & \text{persamaan}\\ & a^{f(x)}=y\end{array}\\ \hline\end{array}$.

$\text{CONTOH SOAL}$.

$\begin{array}{l}\\ 1.& \text{Tentukan himpunan penyelesaian dari}\\ & \text{a}.2^{2x-2021}=1\\ & \text{b}.{\left({\displaystyle \frac{1}{2}}\right)}^{2x-2021}=1\\ & \text{c}.{\sqrt{2}}^{2x-2021}=1\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{|ccc|}\hline \text{a}& \text{b}& \text{c}\\ \begin{array}{rl}{2}^{2x-2021}& =1\\ 2^{2x-2021}& =2^0\\ 2x-2021& =0\\ 2x& =2021\\ x& ={\displaystyle \frac{2021}{2}}\\ & \end{array}& \begin{array}{rl}{\left({\displaystyle \frac{1}{2}}\right)}^{2x-2021}& =1\\ {\left({\displaystyle \frac{1}{2}}\right)}^{2x-2021}& ={\left(\frac{1}{2}\right)}^0\\ 2x-2021& =0\\ 2x& =2021\\ x& ={\displaystyle \frac{2021}{2}}\end{array}& \begin{array}{rl}{\sqrt{2}}^{2x-2021}& =1\\ {\sqrt{2}}^{2x-2021}& ={\sqrt{2}}^0\\ 2x-2021& =0\\ 2x& =2021\\ x& ={\displaystyle \frac{2021}{2}}\\ & \end{array}\\ \mathbf{\text{HP}}=\left\{{\displaystyle \frac{2021}{2}}\right\}& \mathbf{\text{HP}}=\left\{{\displaystyle \frac{2021}{2}}\right\}& \mathbf{\text{HP}}=\left\{{\displaystyle \frac{2021}{2}}\right\}\\ \hline\end{array}\end{array}$.

$\begin{array}{l}\\ 2.& \text{Tentukan himpunan penyelesaian dari}\\ & \text{a}.2^{2x-2021}=128\\ & \text{b}.{\left({\displaystyle \frac{1}{2}}\right)}^{2x-2021}=128\\ & \text{c}.{\sqrt{2}}^{2x-2021}=128\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{|ccc|}\hline \text{a}& \text{b}& \text{c}\\ \begin{array}{rl}{2}^{2x-2021}& =128\\ 2^{2x-2021}& =2^7\\ 2x-2021& =7\\ 2x=7& +2021\\ x={\displaystyle \frac{2028}{2}}& =1014\\ & \end{array}& \begin{array}{rl}{\left({\displaystyle \frac{1}{2}}\right)}^{2x-2021}& =128\\ {\left({\displaystyle \frac{1}{2}}\right)}^{2x-2021}& ={\left({\displaystyle \frac{1}{2}}\right)}^{-7}\\ 2x-2021& =-7\\ 2x=2021& -7\\ x={\displaystyle \frac{2014}{2}}& =1007\end{array}& \begin{array}{rl}{\sqrt{2}}^{2x-2021}& =128\\ {\sqrt{2}}^{2x-2021}& ={\sqrt{2}}^{256}\\ 2x-2021& =256\\ 2x=2021& +256\\ x& ={\displaystyle \frac{2277}{2}}\\ & \end{array}\\ \mathbf{\text{HP}}=\left\{1014\right\}& \mathbf{\text{HP}}=\left\{{\displaystyle 1007}\right\}& \mathbf{\text{HP}}=\left\{{\displaystyle \frac{2277}{2}}\right\}\\ \hline\end{array}\end{array}$.

$\begin{array}{l}\\ 3.& (\mathbf{\text{SPMB 04}})\text{Nilai}x\text{yang memenuhi}\\ & {\displaystyle \frac{27}{3^{2x-1}}={81}^{-0,125}\text{adalah... .}}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}{\displaystyle \frac{27}{3^{2x-1}}}& ={81}^{-0,125}\\ 3^{3-(2x-1)}& =3^{4(\frac{1}{8})}\\ 3-2x+1& =-{\displaystyle \frac{1}{2}}\\ -2x+4& =-{\displaystyle \frac{1}{2}}\\ -x+2& =-{\displaystyle \frac{1}{4}}\\ -x& =-2-{\displaystyle \frac{1}{4}}\\ -x& =-2{\displaystyle \frac{1}{4}}\\ x& =2{\displaystyle \frac{1}{4}}\end{array}\end{array}$.

$\begin{array}{l}\\ 4.& (\mathbf{\text{UMPTN 00}})\\ & \text{Bentuk}{\left(\sqrt[3]{{\displaystyle \frac{1}{243}}}\right)}^{3x}={\left({\displaystyle \frac{3}{3^{x-2}}}\right)}^2\sqrt[3]{{\displaystyle \frac{1}{9}}}\\ & \text{Jika}{x}_0\text{memenuhi persamaan, maka nilai}\\ & 1-{\displaystyle \frac{3}{4}{x}_0=....}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}{\left(\sqrt[3]{{\displaystyle \frac{1}{243}}}\right)}^{3x}& ={\left({\displaystyle \frac{3}{3^{x-2}}}\right)}^2\sqrt[3]{{\displaystyle \frac{1}{9}}}\\ 3^{-5x}& =3^{2(1-(x-2))}{.3}^{-\frac{2}{3}}\\ -5x& =2(1-(x-2))+(-{\displaystyle \frac{2}{3}}),\text{dikali}3\\ -15x& =6(3-x)+(-2)\\ -15x& =18-6x-2\\ 6x-15x& =16\\ -9x& =16\\ x& ={\displaystyle \frac{16}{-9}}\\ x_0& =-{\displaystyle \frac{16}{9},\text{selanjutnya}}\\ 1-{\displaystyle \frac{3}{4}{x}_0}& =1-{\displaystyle \frac{3}{4}\times (-\frac{16}{9})}\\ & =1+\frac{4}{3}\\ & =1+1{\displaystyle \frac{1}{3}}\\ & =2{\displaystyle \frac{1}{3}}\end{array}\end{array}$.

$\begin{array}{l}\\ 5.& \text{Jumlah akar-akar persamaan}\\ & 5^{x+1}+5^{2-x}-30=0\text{adalah}....\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}{5}^{x+1}+5^{2-x}-30& =0\\ \left(5^x\right){.5}^1+{\displaystyle \frac{5^2}{5^x}-30}& =0\\ 5{\left(5^x\right)}^2+25-30\left(5^x\right)& =0\\ \text{Persamaan kuadrat}& \text{dalam}{5}^x,\text{maka}\\ 5(5^x{)}^2-30(5^x)+25& =0\{\begin{array}{ll}a& =5\\ b& =-30\\ c& =25\end{array}\\ (5^{x_1}).\left(5^{x_2}\right)& ={\displaystyle \frac{c}{a}}\\ 5^{x_1+x_2}& ={\displaystyle \frac{25}{5}=5}\\ 5^{x_1+x_2}& =5^1\\ x_1+x_2& =1\end{array}\end{array}$.

$\begin{array}{l}\\ 6.& \text{Jumlah akar-akar persamaan}\\ & {2023}^{x^2-7x+7}={2024}^{x^2-7x+7}\text{adalah}....\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}{2023}^{x^2-7x+7}& ={2024}^{x^2-7x+7}\\ \text{Karena basis}& \text{tidak sama},\\ \text{maka harusl}& \text{ah pangkatnya}=0,\\ x^2-7x+7& =0\\ \text{dan jumlah}& \text{akar-akarnya adalah}:\\ x_1+x_2& =-{\displaystyle \frac{b}{a},\text{dari persamaan}}\\ x^2-7x+7& =0\{\begin{array}{ll}a& =1\\ b& =-7\\ c& =7\end{array}\\ \text{maka}{x}_1+x_2& =-{\displaystyle \frac{b}{a}=-\frac{-7}{1}=7}\end{array}\end{array}$.

$\begin{array}{l}\\ 7.& \text{Tentukan himpunan penyelesaian dari}\\ & (x-2{)}^{x^2-7x+6}=1\text{adalah}....\\ \\ & \mathbf{\text{Jawab}}:\\ & \text{Ingat bentuk}f(x{)}^{g(x)}=1\{\begin{array}{ll}f(x)& =x-2\\ g(x)& =x^2-7x+5\end{array}\\ & \begin{array}{|lll|}\hline f(x)=1& f(x)=-1& g(x)=0\\ & \text{Syarat}g(x)\text{genap}& \text{Syarat}f(x)\ne 0\\ \begin{array}{rl}x& -2=1\\ x& =3\\ & \end{array}& \begin{array}{rl}x& -2=-1\\ x& =2-1=1\\ & \end{array}& \begin{array}{rl}{x}^2& -7x+6=0\\ \iff & (x-1)(x-6)\\ \iff & x=1\text{atau}x=6\end{array}\\ & \begin{array}{rl}\text{S}& \text{yaratnya}x\\ \text{u}& \text{ntuk}x=1\\ g& (1)=1^2-7+6\\ & =0(\text{memenuhi})\end{array}& \begin{array}{rl}f(1)& =1-2=-1\ne 0\\ f(6)& =6-2=4\ne 0\\ & \\ & \end{array}\\ & \mathbf{\text{Catatan}}:0& \\ & \text{paritasnya genap}& \\ \hline\end{array}\\ & \mathbf{\text{HP}}=\{1,3,6\}\end{array}$.

$\begin{array}{l}\\ 8.& \text{Tentukan himpunan penyelesaian dari}\\ & (x^2-9x+19{)}^{2x+3}=(x^2-9x+19{)}^{x-1}\text{adalah}....\\ \\ & \mathbf{\text{Jawab}}:\\ & \text{Ingat bentuk}h(x{)}^{f(x)}=h(x{)}^{g(x)}\{\begin{array}{ll}h(x)& =x^2-9x+19\\ f(x)& =2x+3\\ g(x)& =x-1\end{array}\\ & \begin{array}{|ll|}\hline \begin{array}{rl}& \text{Syarat-syaratnya}\\ & \bullet f(x)=g(x)\\ & \iff 2x+3=x-1\\ & \iff x=-4\\ & \bullet h(x)=1\\ & \iff x^2-9x+19=1\\ & \iff x^2-9x+18=0\\ & \iff (x-3)(x-6)=0\\ & \iff x=3\text{atau}x=6\\ & \bullet h(x)=0\\ & \iff x^2-9x+19=0\\ & \iff x_{1,2}={\displaystyle \frac{9\pm \sqrt{5}}{2}}\\ & \text{gunakan rumus ABC}\\ & \text{Setelah diuji keduanya}\\ & \text{positif, maka}\\ & x={\displaystyle \frac{9\pm \sqrt{5}}{2}\text{merupakan}}\\ & \mathbf{\text{penyelesaian}}\end{array}& \begin{array}{rl}& \text{lanjutannya}\\ & \bullet h(x)=-1\\ & \iff x^2-9x+19=-1\\ & \iff x^2-9x+20=0\\ & \iff (x-4)(x-5)=0\\ & \iff x=4\text{atau}x=5\\ & \text{Uji nilanya}\\ & \text{untuk}x=4\\ & ⧫f(4)=2(4)+3\text{ganjil}\\ & ⧫g(4)=4-1\text{ganjil}\\ & \text{karena}f(4),g(4)\text{keduanya }\\ & \text{ganjil, maka}x=4\\ & \text{adalah}\mathbf{\text{penyelesaian}}\\ & \text{untuk}x=5\\ & ⧫f(5)=2(5)+3\text{ganjil}\\ & ⧫g(5)=5-1\text{genapl}\\ & \text{karena}f(4)\ne g(4),\text{maka}x=5\\ & \text{adalah}\mathbf{\text{bukan penyelesaian}}\\ & \end{array}\\ \hline\end{array}\\ & \mathbf{\text{HP}}=\{-4,3,4,6,{\displaystyle \frac{9-\sqrt{5}}{2},\frac{9+\sqrt{5}}{2}}\}\end{array}$.

DAFTAR PUSTAKA

1. Kurnia, N, dkk. 2016. Jelajah Matematika I SMA Kelas X Peminatan MIPA. Jakarta: YUDHISTIRA.