First slide

Introduction to real valued functions

Question

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

Easy

A

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

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

$2 ab + 1$
D

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

Solution

$ab = \log_{4} 5 . \log_{5} 6 = \log_{4} 6 = \frac{1}{2} \log_{2} 6$
$ab = \frac{1}{2} (1 + \log_{2} 3) \Rightarrow 2 ab - 1 = \log_{2} 3$
$∴ \log_{3} 2 = \frac{1}{2 ab - 1}$ .

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