Question:
Find the value of $x$ for which $\left(\frac{4}{9}\right)^{4} \times\left(\frac{4}{9}\right)^{-7}=\left(\frac{4}{9}\right)^{2 x-1}$
Solution:
Given:
$\left(\frac{4}{9}\right)^{4} \times\left(\frac{4}{9}\right)^{-7}=\left(\frac{4}{9}\right)^{2 x-1}$
$\therefore\left(\frac{4}{9}\right)^{(4-7)}=\left(\frac{4}{9}\right)^{-3}=\left(\frac{4}{9}\right)^{2 x-1}$
$\Rightarrow 2 x-1=-3$
$2 x=-3+1=-2$
$\Rightarrow x=-1$