Question

Find the value of ${\log }_{9}729+{\log }_{2}32$

Answer 1

Let $a={\log }_{9}729+{\log }_{2}32$

$={\log }_9{9}^3+{\log }_2{2}^5$

$=3{\log }_{9}9+5{\log }_{2}2\dots (n{\log }_{a}b={\log }_a{b}^n)$

$=3\times 1+5\times 1$ $(\therefore {\log }_{a}a=1)$

$=8$

Hence, ${\log }_{9}729+{\log }_{2}32=8$