Question

Find the value of $x$ if ${\left(\frac{-3}{2}\right)}^6\times {\left(\frac{4}{9}\right)}^3={\left(\frac{1}{2}\right)}^{3x}$

Answer 1

$\frac{{(-3)}^6}{2^6}\times \frac{4^6}{9^3}={\left(\frac{1}{2}\right)}^{3x}$

$3^6\times 2^{-6}\times {\left(2^2\right)}^3\times {\left(3^2\right)}^{-3}=2^{-3x}$ using $\frac{1}{a^m}=a^{-m}$

$3^{6-6}\times 2^{-6+6}=2^{-3x}$

$\Rightarrow 2^{-3x}=2^0{3}^0$

$\Rightarrow 2^{-3x}=1$ since $a^0=1$

$\Rightarrow 2^{-3x}=2^0$

$\Rightarrow -3x=1$ since bases are same we can equate the powers

$\therefore x=\frac{-1}{3}$