Question

Given ${\left(\frac{8}{27}\right)}^{x-1}={\left(\frac{9}{4}\right)}^{2x+1}$. If the value of $x$ is $\frac{1}{m}$, then $m$ is

Answer 1

${\left(\frac{8}{27}\right)}^{x-1}={\left(\frac{9}{4}\right)}^{2x+1}$

$\Rightarrow {\left(\frac{2}{3}\right)}^{3x-3}={\left(\frac{3}{2}\right)}^{4x+2}$

$\Rightarrow {\left(\frac{2}{3}\right)}^{3x-3}={\left(\frac{2}{3}\right)}^{-4x-2}$

$\Rightarrow 3x-3=-4x-2$

$\Rightarrow 7x=1$

$\Rightarrow x=\frac{1}{7}$

Therefore, $m=7$