$=log_2e - log_4e+log_8e-log_{16}e+log_{32}e...$
$=log_2e-log_{2^2}e+log_{2^3}e-log_{2^4}e...$
$=log_2e-\frac{1}{2}log_{2}e+\frac{1}{3}log_{2}e-\frac{1}{4}log_{2}e..$
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.