Exercise Zone : Persamaan Eksponen

Berikut ini adalah kumpulan soal mengenai Persamaan Eksponen tingkat dasar. Jika ada jawaban yang salah, mohon dikoreksi melalui komentar. Terima kasih.

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No. 1

Nilai $x$ yang memenuhi persamaan ${\dfrac{\sqrt[3]{(0{,}008)^{7-2x}}}{(0{,}2)^{-4x+5}}=1}$ adalah....

1. $-3$
2. $-2$
3. $-1$

4. $0$
5. $1$

$\begin{array}{rl}{\displaystyle \frac{\sqrt[3]{(\mathrm{0,008}{)}^{7-2x}}}{(0,2{)}^{-4x+5}}}& =1\\ {\displaystyle \frac{\sqrt[3]{{((0,2{)}^3)}^{7-2x}}}{(0,2{)}^{-4x+5}}}& =1\\ {\displaystyle \frac{(0,2{)}^{7-2x}}{(0,2{)}^{-4x+5}}}& =1\\ (0,2{)}^{7-2x-(-4x+5)}& =1\\ (0,2{)}^{7-2x+4x-5}& =1\\ (0,2{)}^{2x+2}& =(0,2{)}^0\\ 2x+2& =0\\ 2x& =-2\\ x& =-1\end{array}$

Jadi, $x=-1$.
JAWAB: C

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No. 2

Hasil kali semua $x$ yang memenuhi persamaan ${9^{x^3-4x^2-x+4}-9^{x^2+x-6}=0}$ adalah....

1. $-10$
2. $-5\sqrt2$
3. $5$

4. $5\sqrt2$
5. $10$

$\begin{array}{rl}{9}^{x^3-4x^2-x+4}-9^{x^2+x-6}& =0\\ 9^{x^3-4x^2-x+4}& =9^{x^2+x-6}\\ x^3-4x^2-x+4& =x^2+x-6\\ x^3-5x^2-2x+10& =0\end{array}$
$a=1$, $b=-5$, $c=-2$, $d=10$

$\begin{array}{rl}{x}_1{x}_2{x}_3& =-{\displaystyle \frac{d}{a}}\\ & =-{\displaystyle \frac{10}{1}}\\ & =\boxed{{\displaystyle \boxed{{\displaystyle -10}}}}\end{array}$

Jadi, hasil kali semua $x$ adalah $-10$.
JAWAB: A

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No. 3

Diberikan persamaan: ${\left(3\sqrt{\dfrac1{243}}\right)^{3x}=\left(\dfrac3{3^{x-2}}\right)^2\sqrt[3]{\dfrac19}}$. Jika $x_0$ memenuhi persamaan maka nilai ${1-\dfrac34x_0}$ adalah ....

1. $1\dfrac3{16}$
2. $1\dfrac14$
3. $1\dfrac34$

4. $2\dfrac13$
5. $1\dfrac79$

$\begin{aligned}\left(3\sqrt{\dfrac1{243}}\right)^{3x}&=\left(\dfrac3{3^{x-2}}\right)^2\sqrt[3]{\dfrac19}\\[8pt]\left(3\sqrt{\dfrac1{3^5}}\right)^{3x}&=\left(3^{1-(x-2)}\right)^2\sqrt[3]{\dfrac1{3^2}}\\[8pt]\left(3\sqrt{3^{-5}}\right)^{3x}&=\left(3^{1-x+2}\right)^2\sqrt[3]{3^{-2}}\\\left(3\cdot3^{-\frac52}\right)^{3x}&=\left(3^{-x+3}\right)^2\cdot3^{-\frac23}\\\left(3^{1-\frac52}\right)^{3x}&=3^{-2x+6}\cdot3^{-\frac23}\\3^{-\frac92x}&=3^{-2x+\frac{16}3}\\-\dfrac92x&=-2x+\dfrac{16}3\\-\dfrac92x+2x&=\dfrac{16}3\\-\dfrac52x&=\dfrac{16}3\\x&=-\dfrac{32}{15}\end{aligned}$

$\begin{array}{rl}1-{\displaystyle \frac{3}{4}}{x}_0& =1-{\displaystyle \frac{\cancel{3}}{\cancel{4}}}(-{\displaystyle \frac{{\cancel{32}}^8}{{\cancel{15}}^5}})\\ & =1+{\displaystyle \frac{8}{5}}\\ & =2{\displaystyle \frac{3}{5}}\end{array}$

Jadi, $1-\dfrac34x_0=2\dfrac35$.
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