Question

Find $x$ if ${\left(\frac{-2}{3}\right)}^{-13}\times {\left(\frac{3}{-2}\right)}^8{=\left(\frac{-2}{3}\right)}^{-2x+1}$.

Answer 1

${\left(\frac{-2}{3}\right)}^{-13}\times {\left(\frac{3}{-2}\right)}^8={\left(\frac{-2}{3}\right)}^{-2x+1}$

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

$Since{\left(\frac{a}{b}\right)}^m={\left(\frac{b}{a}\right)}^{-m}$

${\left(\frac{3}{-2}\right)}^{13+8}={\left(\frac{3}{-2}\right)}^{2x-1}$

${\left(\frac{3}{-2}\right)}^{21}={\left(\frac{3}{-2}\right)}^{2x-1}$

$Since{a}^m=a^n\Rightarrow m=n$

$2x-1=21$

$2x=22$

$x=11$