First slide

Types of real valued functions XI

Question

If $\log_{4} 5 = a and \log_{5} 6 = b, then \log_{3} 2$ is equal to

Moderate

A

$\frac{1}{2 a + 1}$
B

$\frac{1}{2 b + 1}$
C

$2 ab + 1$
D

$\frac{1}{2 ab - 1}$

Solution

$Here. 5 = 4^{a} and 6 = 5^{b} .$
$Let \log_{3} 2 = x, then 2 = 3^{x} .$
$now 6 = 5^{b} = {(4^{a})}^{b} = 4^{ab}$
$\begin{array}{l} ∴ 2 = {(2^{2 ab - 1})}^{x} = 2^{x (2 ab - 1)} \\ \Rightarrow x (2 ab - 1) = 1 \end{array}$

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