Question
Find $x$, if : $\left(\sqrt[3]{\frac{2}{3}}\right)^{x-1}=\frac{27}{8}$
✓
Answer
$\left(\sqrt[3]{\frac{2}{3}}\right)^{x-1}=\frac{27}{8}$
$ {\left[\left(\frac{2}{3}\right)^{\frac{1}{3}}\right]^{x-1}=\frac{3^3}{2^3}}$
$ \Rightarrow\left(\frac{2}{3}\right)^{\frac{x-1}{3}}=\left(\frac{3}{2}\right)^3$
$ \Rightarrow\left(\frac{2}{3}\right)^{\frac{x-1}{3}}=\left(\frac{2}{3}\right)^{-3}$
We know that if bases are equal, the powers are equal
$\Rightarrow \frac{x-1}{3}=-3$
$ \Rightarrow x-1=-9$
$ \Rightarrow x=-9+1$
$ \Rightarrow x=-8$
$ {\left[\left(\frac{2}{3}\right)^{\frac{1}{3}}\right]^{x-1}=\frac{3^3}{2^3}}$
$ \Rightarrow\left(\frac{2}{3}\right)^{\frac{x-1}{3}}=\left(\frac{3}{2}\right)^3$
$ \Rightarrow\left(\frac{2}{3}\right)^{\frac{x-1}{3}}=\left(\frac{2}{3}\right)^{-3}$
We know that if bases are equal, the powers are equal
$\Rightarrow \frac{x-1}{3}=-3$
$ \Rightarrow x-1=-9$
$ \Rightarrow x=-9+1$
$ \Rightarrow x=-8$
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