Question

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

• A. $3.6$
• B. $5$
• C. $4$
• D. None of these

Answer 1

The correct option is A

$3.6$

Explanation for the correct option:

Solving the given expression:.

$$\begin{gathered} 32\sqrt[5]{4}=2^5\times 2^{\frac{2}{5}} \\ =2^{\frac{27}{5}} \end{gathered}$$

$2\sqrt{2}=2^{\frac{3}{2}}$

Now,

$$\begin{gathered} {\log }_{2^{\frac{3}{2}}}{2}^{\frac{27}{5}}=\frac{27}{5}{\log }_{2^{\frac{3}{2}}}2 \\ =\left(\frac{27}{5}\right)\left(\frac{2}{3}\right){\log }_{2}2 \\ =\frac{18}{5} \\ =3.6 \end{gathered}$$ [ $\mathbf{\because }{\mathbf{log}}_{{\mathbf{b}}^{\mathbf{n}}}\mathbf{a}\mathbf{=}\frac{\mathbf{1}}{\mathbf{n}}{\mathbf{log}}_{\mathbf{b}}\mathbf{a}$,${\mathbf{log}}_{\mathbf{b}}{\mathbf{a}}^{\mathbf{n}}\mathbf{=}{\mathbf{nlog}}_{\mathbf{b}}\mathbf{a}$]

Hence, the correct answer is option (A)