If $y=1-x+\frac{x^2}{2!}-\frac{x^3}{3!}+\dots$ and $z=-y-\frac{y^2}{2}-\frac{y^3}{3}-\dots$ then $\ln (\frac{1}{1-e^x})$

Question

For a nonzero number $x$, if $y=1-x+\frac{x^2}{2!}-\frac{x^3}{3!}+\dots$ and $z=-y-\frac{y^2}{2}-\frac{y^3}{3}-\dots$ then the value of $\ln (\frac{1}{1-e^z})$ is .....

I can see that $y=e^{-x}$ and $z=\ln {(1-y)}$. And, so $z=\ln(1-e^{-x})$. But how to go further?

What is the exact question? Do you want an answer in only $y$, or only $z$, or both? — Marc, Apr 10 '14 at 14:26
@Marc, sorry. I have edited the question. The answer is $x$. — Silent, Apr 10 '14 at 14:29

score 1 · Accepted Answer · answered Apr 10 '14 at 14:38

Once you know that $z=\log(1-e^{-x})$, you immediately have that $$\log\left(\frac{1}{1-e^{z}}\right)=\log\left(\frac{1}{1-e^{\log(1-e^{-x})}}\right)= \log\left(\frac{1}{1-(1-e^{-x})}\right)= \log\left(\frac{1}{e^{-x}}\right)=\log e^x=x $$ And you've done.

If $y=1-x+\frac{x^2}{2!}-\frac{x^3}{3!}+\dots$ and $z=-y-\frac{y^2}{2}-\frac{y^3}{3}-\dots$ then $\ln (\frac{1}{1-e^x})$

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