Question

Find the value of $-x$, so that-

${\left(\frac{3}{4}\right)}^{-9}\times {\left(\frac{3}{4}\right)}^{-7}={\left(\frac{3}{4}\right)}^{4x}$

Answer 1

Given that, ${\left(\frac{3}{4}\right)}^{-9}\times {\left(\frac{3}{4}\right)}^{-7}={\left(\frac{3}{4}\right)}^{4x}$

$\Rightarrow {\left(\frac{3}{4}\right)}^{-9+(-7)}={\left(\frac{3}{4}\right)}^{4x}\left(since{x}^a\times x^b=x^{a+b}\right)$

On equating the powers,

$\Rightarrow -9+(-7)=4x$

$\Rightarrow -9-7=4x$

$\Rightarrow 4x=-16$

$\Rightarrow x=\frac{-16}{4}$

$\Rightarrow x=-4$

$\therefore -x=4$