# If y=1−x+x^2/(2!)−x^3/(3!)+… and z=−y−y^2/2−y^3/3−… then ln(1/(1−e^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 $\mathrm{ln}\left(\frac{1}{1-{e}^{x}}\right)$
I can see that $y={e}^{-x}$ and $z=\mathrm{ln}\left(1-y\right)$. And, so $z=\mathrm{ln}\left(1-{e}^{-x}\right)$. But how to go further?
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Ben Logan
Once you know that $z=\mathrm{log}\left(1-{e}^{-x}\right)$, you immediately have that
$\mathrm{log}\left(\frac{1}{1-{e}^{z}}\right)=\mathrm{log}\left(\frac{1}{1-{e}^{\mathrm{log}\left(1-{e}^{-x}\right)}}\right)=\mathrm{log}\left(\frac{1}{1-\left(1-{e}^{-x}\right)}\right)=\mathrm{log}\left(\frac{1}{{e}^{-x}}\right)=\mathrm{log}{e}^{x}=x$
And you've done.