Q42 / 110 · VITEEE 2024 · Mathematics
$\log \left(\log _{a b} a+\frac{1}{\log _b a b}\right)$ is (where $\left.a b \neq 1\right)$
- Correct answerA0
- B1
- C$\log _a a b$
- D$\log _b a b$
Solution
$\begin{aligned} & \log \left(\log _{a b} a+\frac{1}{\log _b a b}\right) \\ & =\log \left(\log _{a b} a+\log _{a b} b\right)\left\{\because \log _m n=\frac{1}{\log _n m}\right\} \\ & =\log \left(\log _{a b}(a b)\right)\end{aligned}$
$\left\{\because \log _b m+\log _b n=\log _b(m n)\right\}$
$=\log (1)=0$
Hence, the correct answer is option: 'A'.