log _{5}left(1+frac{1}{5}right)+log _{5}left( | vedclass

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(C) The given expression is $\log _{5}\left(1+\frac{1}{5}\right)+\log _{5}\left(1+\frac{1}{6}\right)+\log _{5}\left(1+\frac{1}{7}\right)+\cdots+\log _

Vedclass · Accepted Answer

$3$

Vedclass · Answer

(C) The given expression is $\log _{5}\left(1+\frac{1}{5}\right)+\log _{5}\left(1+\frac{1}{6}\right)+\log _{5}\left(1+\frac{1}{7}\right)+\cdots+\log _{5} \left(1+\frac{1}{624}\right)$.Simplify each term inside the logarithm: $\log _{5} \left(\frac{6}{5}\right)+\log _{5} \left(\frac{7}{6}\right)+\log _{5} \left(\frac{8}{7}\right)+\cdots+\log _{5} \left(\frac{625}{624}\right)$.Using the property $\log_{b}(x) + \log_{b}(y) = \log_{b}(xy)$,we can combine the terms:$\log _{5} \left(\frac{6}{5} \cdot \frac{7}{6} \cdot \frac{8}{7} \cdot \cdots \cdot \frac{625}{624}\right)$.Observe that the terms cancel out in a telescoping manner:$\log _{5} \left(\frac{625}{5}\right)$.Calculate the division: $\frac{625}{5} = 125$.So,the expression becomes $\log _{5}(125)$.Since $125 = 5^3$,we have $\log _{5}(5^3) = 3 \log _{5}(5) = 3(1) = 3$.

$\log _{5}\left(1+\frac{1}{5}\right)+\log _{5}\left(1+\frac{1}{6}\right)+\log _{5}\left(1+\frac{1}{7}\right)+\cdots+\log _{5} \left(1+\frac{1}{624}\right)$

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