If log 7 2- m- then log 49 28 is equal toA- -zn-2B- 2-1-2 m- 21-2 m D -1- m

The correct option is A 1+2m/2log_49 28=log 28/log 49=log 7+ log 4/2 log 7 =log 7/2 log 7+log 4/2 log 7=1/2+1/2log_7 4 =1/2+1/2.2 log_7 2=1/2+log_7 2=1/2+m=1+2 ...

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Byju's Answer

Standard XI

Mathematics

Fundamental Laws of Logarithms

If log 7 2= m...

Question

If $l o g_{7} 2 = m, t h e n l o g_{49} 28$ is equal to

2 (1 + 2 m)

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\frac{1 + 2 m}{2}

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\frac{2}{1 + 2 m}

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1 + m

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Solution

The correct option is B $\frac{1 + 2 m}{2}$
$l o g_{49} 28 = \frac{l o g 28}{l o g 49} = \frac{l o g 7 + l o g 4}{2 l o g 7}$
$= \frac{l o g 7}{2 l o g 7} + \frac{l o g 4}{2 l o g 7} = \frac{1}{2} + \frac{1}{2} l o g_{7} 4$
$= \frac{1}{2} + \frac{1}{2} .2 l o g_{7} 2 = \frac{1}{2} + l o g_{7} 2 = \frac{1}{2} + m = \frac{1 + 2 m}{2}$

Suggest Corrections

If log72=m, then log4928 is equal to

The correct option is B 1+2m2log49 28=log 28log 49=log 7+log 42 log 7 =log 72 log 7+log 42 log 7=12+12log7 4 =12+12.2 log7 2=12+log7 2=12+m=1+2m2

If $l o g_{7} 2 = m, t h e n l o g_{49} 28$ is equal to