Question

The number $N=\left(\operatorname{antilog}_5\left(\log _{\sqrt{5}}(14)\right)\right) \cdot\left(\log _9\left(\operatorname{antilog}_3\left(2^{-1}\right)\right)\right)$, then $\sqrt{ N }$ is relatively prime with

• A. 30
• B. 42
• C. 70
• D. 105

Answer 1

Answer: (A)

$ N =\left(5^{\log _{\sqrt{5}} 14}\right)\left(\log _9 3^{2^{-1}}\right)=(14)^2 \times \frac{2^{-1}}{2}=49 $
$\Rightarrow \sqrt{ N }=7 $