Question

If $x=1+\frac{1}{2\times 1!}+\frac{1}{4\times 2!}+\frac{1}{8\times 3!}$ and $y=1+\frac{x^2}{1!}+\frac{x^4}{2!}+\frac{x^6}{3!}+.....$
Then, the value of ${\log }_{e}y$ is

• A. $e$
• B. $e^2$
• C. $1$
• D. $\frac{1}{e}$

Answer 1

The correct option is A $e$

Given $x=1+\frac{1}{2\times 1!}+\frac{1}{4\times 2!}+\frac{1}{8\times 3!}$
$\Rightarrow x=1+\frac{{\left(1/2\right)}^1}{1!}+\frac{{\left(1/2\right)}^2}{2!}+\frac{{\left(1/2\right)}^3}{3!}+...$
$\Rightarrow x=e^{1/2}\Rightarrow x^2=e....\left(i\right)$
and
$y=1+\frac{x^2}{1!}+\frac{x^4}{2!}+\frac{x^6}{3!}+..$
$y=1+\frac{{\left(x^2\right)}^1}{1!}+\frac{{\left(x^2\right)}^2}{2!}+\frac{{\left(x^2\right)}^3}{3!}+..$
$\Rightarrow y=e^{x^2}=e^e$ (from eq.$\left(i\right)$)
Taking $\log$ on both sides, we get
${\log }_{e}y=e$