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  4. If log 105 7= a , log 7 5= b then log 35 105 is equal to

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Q. If $\log _{105} 7= a , \log _7 5= b$ then $\log _{35} 105$ is equal to

Continuity and Differentiability

Solution:

$a b=\log _{105} 7 \cdot \log _7 5=\log _{105} 5$
Now $\log _{35} 105=\frac{1}{\log _{105} 35}=\frac{1}{\log _{105} 5+\log _{105} 7}=\frac{1}{a b+a}=\frac{1}{a(b+1)}$