MCQ
If ${\log _4}5 = a$ and ${\log _5}6 = b,$ then ${\log _3}2$ is equal to
- A${1 \over {2a + 1}}$
- B${1 \over {2b + 1}}$
- C$2ab + 1$
- ✓${1 \over {2ab - 1}}$
$ab = {1 \over 2}(1 + {\log _2}3) \Rightarrow 2ab - 1 = {\log _2}3$
$\therefore {\log _3}2 = {1 \over {2ab - 1}}$.
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