Question

If ${\left(\sqrt[3]{3\frac{2}{3}}\right)}^{x-1}=\frac{27}{8}$, then the value of $x+16$ is equal to:

Answer 1

$\Rightarrow {\left(\sqrt[3]{3\frac{2}{3}}\right)}^{x-1}=\frac{27}{8}$

$\Rightarrow {\left({\left(\frac{2}{3}\right)}^{\frac{1}{3}}\right)}^{x-1}=\frac{3^3}{2^3}$

$\Rightarrow {\left(\frac{2}{3}\right)}^{\frac{x}{3}-\frac{1}{3}}={\left(\frac{2}{3}\right)}^{-3}$

Both side base are same so
$\Rightarrow \frac{x}{3}-\frac{1}{3}=-3$

$\Rightarrow x-1=-9$

$\Rightarrow x=-8$

$x+16=-8+16=8$