Question

The value of $1+\frac{x{\log }_{e}2}{1!}+\frac{x^2}{2!}({\log }_{e}2{)}^2+\frac{x^3}{3!}({\log }_{e}2{)}^3+...\infty$s equal to

• A. $e^2$
• B. $a^2$
• C. $2^a$
• D. $2^x$

Answer 1

The correct option is D $2^x$

Series expansion of $e^x$ is

$e^x=1+\frac{x}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+........\infty$

So,

$1+\frac{x{\log }_{e}2}{1!}+\frac{{\left(x{\log }_{e}2\right)}^2}{2!}+\frac{{\left(x{\log }_{e}2\right)}^3}{3!}+........\infty =e^{x{\log }_{e}2}$

$=e^{{\log }_e{2}^x}$

$={\left(2^x\right)}^{{\log }_{e}e}$ $(a^{\log b}=b^{\log a})$

$=2^x$