If log72 = m, then log4928 is equal to ? 1/(1+2m) (1+2m)/2 2m/(2m+1) (2m+1)/2m l

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Question
If log72 = m, then log4928 is equal to ?
Options
A. 1/(1+2m)
B. (1+2m)/2
C. 2m/(2m+1)
D. (2m+1)/2m

Correct Answer

(1+2m)/2

Explanation

$\log_{49} (28) = \frac{1}{2 \log_{7} (7 � 4)}$

= $\frac{1}{2} + \frac{1}{2} (2 \log_{7} (2))$
= $\frac{1}{2} + \log_{7} (2)$
= $\frac{1}{2} + m$
= $\frac{1 + 2 m}{2}$ .

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