Question

${\left(\frac{4}{9}\right)}^x\times {\left(\frac{3}{2}\right)}^{-1}=\frac{8}{27}$ then x =

Answer 1

${\left(\frac{4}{9}\right)}^x\times {\left(\frac{3}{2}\right)}^{-1}=\frac{8}{27}$

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

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

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

Since bases are same we can equate the powers,

$\Rightarrow 2x+1=3$

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

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