Question

Find the value of '$x$' such that
$\frac{1}{49}\times 7^{2x}=7^8$

Answer 1

$\frac{1}{49}\times 7^{2x}=7^8$
$\frac{1}{7^2}\times 7^{2x}=7^8$ $(\because a^{-m}=\frac{1}{a^m})$
$7^{-2}\times 7^{2x}=7^8$
$7^{2x-2}=7^8$
As bases are equal, Hence
$2x-2=8$
$2x=8+10$
$2x=10$
$x=\frac{10}{2}=5$
$\therefore$ $x=5$