Question

Let $N=\left(\frac{4^5+4^5+4^5+4^5}{3^5+3^5+3^5}\right)\cdot \left(\frac{6^5+6^5+6^5+6^5+6^5+6^5}{2^5+2^5}\right)$, then the value of ${\log }_{2}N$ is equal to

• A. 10
• B. 11
• C. 12
• D. 14

Answer 1

Answer: (C)

$N=\frac{4^6}{3^6}\cdot \frac{6^6}{2^6}=4^6=2^{12}$
$\therefore {\log }_{2}N={\log }_2{2}^{12}=12$