Question

$1+\log x+\frac{(\log x{)}^2}{2!}+\frac{{(\log x)}^3}{3!}+....\infty$is

• A. $-\log x$
• B. $\log x$
• C. $2\log x$
• D. $x$

Answer 1

The correct option is D

$x$

Explanation for the correct option:

Step 1. Use expansion formula of $e^x$

To find $1+\log x+\frac{(\log x{)}^2}{2!}+\frac{{(\log x)}^3}{3!}+....\infty$

As we know, ${\mathit{e}}^{\mathbf{x}}\mathbf{}\mathbf{=}\mathbf{}\mathbf{1}\mathbf{}\mathbf{+}\frac{\mathbf{}\mathbf{x}}{\mathbf{1}\mathbf{!}}\mathbf{+}\mathbf{}\frac{{\mathbf{x}}^{\mathbf{2}}}{\mathbf{2}\mathbf{!}}\mathbf{}\mathbf{+}\mathbf{}\frac{{\mathbf{x}}^{\mathbf{3}}}{\mathbf{3}\mathbf{!}}\mathbf{}\mathbf{+}\mathbf{\dots }\mathbf{.}\mathbf{.}\mathbf{\infty }$

Step 2. Put $x=\log x$, in above expansion

$\Rightarrow$ $e^{\log x}=1+\log x+\frac{(\log x{)}^2}{2!}+\frac{{(\log x)}^3}{3!}+\dots .\infty$

Also, $e^{\log x}=\mathbf{}x$ $\left[\mathbf{\because }{\mathit{e}}^{\mathbf{l}\mathbf{o}\mathbf{g}\mathbf{}\mathbf{a}}\mathbf{=}\mathit{a}\right]$

$\mathbf{\therefore }\mathbf{1}\mathbf{+}\mathit{l}\mathit{o}\mathit{g}\mathbf{}\mathit{x}\mathbf{}\mathbf{+}\frac{\mathbf{(}\mathbf{l}\mathbf{o}\mathbf{g}\mathbf{}\mathbf{x}{\mathbf{)}}^{\mathbf{2}}}{\mathbf{2}\mathbf{!}}\mathbf{}\mathbf{+}\mathbf{}\frac{{\mathbf{(}\mathbf{l}\mathbf{o}\mathbf{g}\mathbf{}\mathbf{x}\mathbf{)}}^{\mathbf{3}}}{\mathbf{3}\mathbf{!}}\mathbf{}\mathbf{+}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{\infty }\mathbf{=}\mathit{x}$

Hence, Option ‘D’ is Correct.