Question

If ${\log }_{2\sqrt{2}}(32\sqrt[5]{4})=\alpha +\beta$ where $\alpha$ is an integer and $\beta \in [0,1)$, then $\alpha$ is

• A. 3
• B. 4
• C. 5
• D. 6

Answer 1

Answer: (A)

Given, ${\log }_{2^{\frac{3}{2}}}\left(2^5\cdot 2^{\frac{2}{5}}\right)=\alpha +\beta ={\log }_{2^{\frac{3}{2}}}\left(2^{\frac{27}{5}}\right)=\frac{27/5}{3/2}=\frac{27}{5}\times \frac{2}{3}=\frac{18}{5}=3.6$
$\therefore \alpha =3,\beta =0.6$