The value of log_{3} e - log_{9} e + log_{27} | vedclass

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(A) Let the given series be $S = \log_{3} e - \log_{9} e + \log_{27} e - \dots$Using the property $\log_{a^n} b = \frac{1}{n} \log_{a} b$,we can write

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$\log_{3} 2$

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(A) Let the given series be $S = \log_{3} e - \log_{9} e + \log_{27} e - \dots$Using the property $\log_{a^n} b = \frac{1}{n} \log_{a} b$,we can write:$S = \log_{3} e - \frac{1}{2} \log_{3} e + \frac{1}{3} \log_{3} e - \dots$Factor out $\log_{3} e$:$S = (\log_{3} e) \left( 1 - \frac{1}{2} + \frac{1}{3} - \dots \right)$We know the logarithmic series expansion $\ln(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \dots$,so for $x=1$,$\ln(2) = 1 - \frac{1}{2} + \frac{1}{3} - \dots$Thus,$S = (\log_{3} e) \ln(2) = \frac{\ln e}{\ln 3} \cdot \ln 2 = \frac{1}{\ln 3} \cdot \ln 2 = \log_{3} 2$.

The value of $\log_{3} e - \log_{9} e + \log_{27} e - \dots$ is equal to

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