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Question: If \(\log_{7}2 = m,\) then \(\log_{49}28\)is equal to...

If log⁡72=m,\log_{7}2 = m, then log⁡4928\log_{49}28is equal to

A

2(1+2m)2(1 + 2m)

B

1+2m2\frac{1 + 2m}{2}

C

21+2m\frac{2}{1 + 2m}

D

1+m1 + m

Answer

1+2m2\frac{1 + 2m}{2}

Explanation

Solution

log⁡4928=log⁡28log⁡49=log⁡7+log⁡42log⁡7\log_{49}28 = \frac{\log 28}{\log 49} = \frac{\log 7 + \log 4}{2\log 7}

=log⁡72log⁡7+log⁡42log⁡7=12+12log⁡74\frac{\log 7}{2\log 7} + \frac{\log 4}{2\log 7} = \frac{1}{2} + \frac{1}{2}\log_{7}4=12+12.2log⁡72\frac{1}{2} + \frac{1}{2}.2\log_{7}2

=12+log⁡72=12+m=1+2m2\frac{1}{2} + \log_{7}2 = \frac{1}{2} + m = \frac{1 + 2m}{2}