Question

Compute the following
${\left(\frac{1}{49}\right)}^{1+{\log }_{7}2}+5^{-{\log }_{1/5}7}$

Answer 1

$={\left(\frac{1}{49}\right)}^{1+lo{g}_7^2}+5^{-lo{g}_{1/5}^7}$

$={\left(\frac{1}{49}\right)}^{lo{g}_7^{14}}+7^{lo{g}_{1/5}^{1/5}}$

$={14}^{lo{g}_7^{1/49}}+7$

$=(14{)}^{-2}+7$

$={\left(\frac{1}{14}\right)}^2+7$.