Question

Evaluate:

$e^{(x-1)-\frac{{(x-1)}^2}{2}+\frac{{(x-1)}^3}{3}-\frac{{(x-1)}^4}{4}+...\infty }$

• A. $\log (x-1)$
• B. $\log x$
• C. $x$
• D. None of these

Answer 1

Answer: (C)

$x$

we know that,

$\log (1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+........$

Replacing $x=x-1$, we get

$\log \left(x\right)=(x-1)-\frac{(x-1{)}^2}{2}+\frac{(x-1{)}^3}{3}-\frac{(x-1{)}^4}{4}+........$

Now,

$e^{(x-1)-\frac{(x-1{)}^2}{2}+\frac{(x-1{)}^3}{3}-\frac{(x-1{)}^4}{4}+........}=e^{{\log }_{e}x}=x$

Therefore, Correct Answer is $x$