ISI2015-DCG-23

Question

The value of $\log _2 e – \log _4 e + \log _8 e – \log _{16} e + \log_{32} e – \cdots$ is

• A. $-1$
• B. $0$
• C. $1$
• D. None of these

Answer 1

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

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

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

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.

Answer 2

And=1 using $log_{e}(1+x)=x-x^{2}/2+.........$

Apply formule $\log_{b^{n}}a=1/n*log_{b} a$

 

Comments

What is the answer of this question??