Question

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

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

Answer 1

Answer: (D)

$2^x$

We know that

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

Replace $x$ by $x{\log }_{e}2$ in above expansion, we get

$e^{x{\log }_{e}2}=1+\frac{x{\log }_{e}2}{1!}+\frac{x^2}{2!}({\log }_{e}2{)}^2+\cdots \infty$

Therefore,

The sum is $e^{x{\log }_{e}2}$

Which is equivalent to $2^x$.