What is the answer of this question??

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gatecse
asked
in Quantitative Aptitude
Sep 18, 2019
recategorized
Nov 14, 2019
by Lakshman Patel RJIT

247 views
1 vote

$=log_2e - log_4e+log_8e-log_{16}e+log_{32}e...$

$=log_2e-log_{2^2}e+log_{2^3}e-log_{2^4}e...$

$=log_2e-\frac{1}{2}log_{2}e+\frac{1}{3}log_{2}e-\frac{1}{4}log_{2}e..$

On taking out $log_2e$ common, we get$:$

$=log_2e(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...) $$ ---> (1)$

Now, since $log(1+x) = x = \frac{x^2}{2} + \frac{x^3}{3}-...$ $--->(2)$

Here, $x = 1$

From $(1)$ and $(2)$, we get $:$

$log_e2 * log_2e$ = $1$

Therefore option $(c)$ is the correct answer.

$=log_2e-log_{2^2}e+log_{2^3}e-log_{2^4}e...$

$=log_2e-\frac{1}{2}log_{2}e+\frac{1}{3}log_{2}e-\frac{1}{4}log_{2}e..$

On taking out $log_2e$ common, we get$:$

$=log_2e(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...) $$ ---> (1)$

Now, since $log(1+x) = x = \frac{x^2}{2} + \frac{x^3}{3}-...$ $--->(2)$

Here, $x = 1$

From $(1)$ and $(2)$, we get $:$

$log_e2 * log_2e$ = $1$

Therefore option $(c)$ is the correct answer.