Question

Logarithm of $32 \sqrt[5]{4}$ to the base $2 \sqrt{2}$ is

Easy

Solution

Let x be the required logarithm, then by definition
${(2 \sqrt{2})}^{x} = 32 \sqrt[5]{4} \Rightarrow {(2 . 2^{1 / 2})}^{x} = 2^{5} . 2^{2 / 5}; ∴ 2^{\frac{3 x}{2}} = 2^{5 + \frac{2}{5}}$
Here, by equating the indices, $\frac{3}{2} x = \frac{27}{5}$
$∴ x = \frac{18}{5} = 3.6 .$

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

SCAN ME