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

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

A

1+2k4\frac{1 + 2k}{4}

B

1+2k2\frac{1 + 2k}{2}

C

1+2k3\frac{1 + 2k}{3}

D

None of these

Answer

1+2k2\frac{1 + 2k}{2}

Explanation

Solution

log⁡4928\log_{49}28= 12\frac{1}{2}log7 (7 × 4) = 12\frac{1}{2} [1+2log⁡721 + 2\log_{7}2]

= 12\frac{1}{2} (1+2k) =1+2k2\frac{1 + 2k}{2}