Question

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

• A. 3.6
• B. 5
• C. 5.6
• D. None of these

Answer 1

Answer: (A)

3.6

Answer 2

Let x be the required logarithm , then by definition

$\mathbf{(2}\sqrt{\mathbf{2}}\mathbf{)}^{\mathbf{x}}\mathbf{= 32}\sqrt[\mathbf{5}]{\mathbf{4}}\mathbf{(2.}\mathbf{2}^{\mathbf{1/2}}\mathbf{)}^{\mathbf{x}}\mathbf{=}\mathbf{2}^{\mathbf{5}}\mathbf{.}\mathbf{2}^{\mathbf{2/5}}$; ∴ $\mathbf{2}^{\frac{\mathbf{3x}}{\mathbf{2}}}\mathbf{=}\mathbf{2}^{\mathbf{5 +}\frac{\mathbf{2}}{\mathbf{5}}}$

Here, by equating the indices, $\frac{3}{2}x = \frac{27}{5}$ , $\therefore x = \frac{18}{5} = 3.6$