If $\log_{4} (5) = a a n d \log_{5} (6) = b$ then $\log_{3} (2)$ is equal to:

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Solution

Find the product of $a$ and $b$ :
Given $\log_{4} (5) = a . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (1) \log_{5} (6) = b . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (2)$
$a b = \log_{4} (5) \times \log_{5} (6) [M u l t i p l y i n g e q u a t i o n (1) a n d (2)] \Rightarrow a b = \frac{\log (5)}{\log (4)} \times \frac{\log (6)}{\log (5)} [∵ l o g_{a} (b) = \frac{\log (b)}{\log (a)}] \Rightarrow a b = \frac{\log (6)}{\log (4)} \Rightarrow a b = \frac{\log (3 \times 2)}{\log (2^{2})} \Rightarrow a b = \frac{\log (3) + \log (2)}{2 \log (2)} [∵ l o g (ab) = l o g (a) + l o g (b)] \Rightarrow 2 a b = \frac{\log (2)}{\log (2)} + \frac{\log (3)}{\log (2)} \Rightarrow 2 a b = 1 + \log_{2} (3) \Rightarrow \log_{2} (3) = 2 a b - 1 \Rightarrow \log_{3} (2) = \frac{1}{2 a b - 1} [∵ l o g_{a} (b) = \frac{1}{\log_{b} (a)}]$
Final answer:
Hence, $l o g_{3} (2)$ $= \frac{1}{2 a b - 1}$

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