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Q. If $\log_{7} \,2 = λ$, then the value of $\log_{49}\, \left(28\right)$ is

WBJEEWBJEE 2011

Solution:

Given, $\log _{7} 2=\lambda$
$\therefore \log _{49}(28) =\log _{7^{2}}\left(2^{2} \times 7\right) $
$=\frac{1}{2}\left[2 \log _{7} 2+\log _{7} 7\right] $
$=\frac{1}{2}(2 \lambda+1) $