Question

Find $x$
${\left[{\left\{{\left(\frac{2}{5}\right)}^2\right\}}^4\right]}^{x+2}={\left[{\left\{{\left(\frac{2}{5}\right)}^{-2}\right\}}^{(x-1)}\right]}^{-3}$

Answer 1

We have,

${\left[{\left\{{\left(\frac{2}{5}\right)}^2\right\}}^4\right]}^{x+2}={\left[{\left\{{\left(\frac{2}{5}\right)}^{-2}\right\}}^{(x-1)}\right]}^{-3}$

${\left\{{\left(\frac{2}{5}\right)}^2\right\}}^{4x+8}={\left\{{\left(\frac{2}{5}\right)}^{-2}\right\}}^{(-3x+3)}$

${\left\{\frac{2}{5}\right\}}^{8x+16}={\left\{\frac{2}{5}\right\}}^{(6x-6)}$

On comparing both sides, we get

$8x+16=6x-6$

$8x-6x=-16-6$

$2x=-22$

$x=-11$

Hence, this is the answer.