Question

The logarithm of $32\sqrt[5]{54}$ to the base $2\sqrt{2}$ is equal to

• A. $3.6$
• B. $3$
• C. $4$
• D. $5$

Answer 1

The correct option is A $3.6$

Consider,

${\log }_{2\sqrt{2}}32\sqrt[5]{54}={\log }_{\left(2^{\frac{3}{2}}\right)}\left(2^5{4}^{\frac{1}{5}}\right)$

$={\log }_{\left(2^{\frac{3}{2}}\right)}\left(2^{5+\frac{2}{5}}\right)$

$={\log }_{\left(2^{\frac{3}{2}}\right)}\left(2^{\frac{27}{5}}\right)$

We know that,

${\log }_{y^a}x=\frac{1}{a}{\log }_{y}x$

$({\log }_y{x}^n=n{\log }_{y}x)$

$=\frac{2}{3}\frac{27}{5}{\log }_{2}2[\because {\log }_{a}a=1]$

$=\frac{18}{5}=3.6$