Question

Find $x$ such that
${\left[{\left(\frac{-2}{7}\right)}^2\right]}^x\times {\left(\frac{-7}{2}\right)}^{-1}=\frac{-8}{343}$ Hence, find ${\left(2x\right)}^2$

Answer 1

$\left[{\left(\frac{-2}{7}\right)}^{2x}\right]\times {\left(\frac{-7}{2}\right)}^{-1}=\frac{-8}{343}$ since ${\left(a^m\right)}^n=a^{mn}$

$\Rightarrow \left[{\left(\frac{-2}{7}\right)}^{2x}\right]\times \left(\frac{-2}{7}\right)=\frac{-8}{343}$ since $\frac{1}{a^m}=a^{-m}$

$\Rightarrow {\left(\frac{-2}{7}\right)}^{2x+1}=\frac{-2^3}{7^3}$

$\Rightarrow {\left(\frac{-2}{7}\right)}^{2x+1}={\left(\frac{-2}{7}\right)}^3$

Since bases are same, we can equate powers

$\Rightarrow 2x+1=3$

$\Rightarrow 2x=3-1=2$

$\Rightarrow x=\frac{2}{2}=1$

$\therefore x=1$

${\left(2x\right)}^2={(2\times 1)}^2=4$