# How can solving for $u$ to obtain the inverse function be completed here?

The text I'm working in has this cdf function

$$F(x)=1-\left(1+5 e^{-x}\right)^{-.2}, \quad-\infty

For the inverse of the cdf, it then sets $$u=F(x)$$ and solves for $$u$$. After that it provides this for the inverse cdf without showing any steps.

$$F^{-1}(u)=\log \left\{.2\left[(1-u)^{-5}-1\right]\right\}, \quad 0

I am trying to work through and solve this on my own and I am getting stuck at what should be the last step. I'm coming up with the inverse function they provide, but it's equal to $$-x$$, not $$x$$. Is there some obvious way to get to just $$x$$ from where I'm leaving off, or am I making some error working my way to it?

\begin{align} u &= 1-\left(1+5 e^{-x}\right)^{-.2} \\ 1-u &= \left(1+5 e^{-x}\right)^{-.2} \\ (1-u)^{-5} &= 1+5 e^{-x} \\ (1-u)^{-5} - 1 &= 5 e^{-x} \\ .2[(1-u)^{-5} - 1] &= e^{-x} \\ \log \left\{.2\left[(1-u)^{-5}-1\right]\right\}&=-x \\ \end{align}

At this point $$F^{-1}(u)$$ should equal $$x$$, but I'm stuck with $$-x$$. Thank you for any assistance!

Your derivation is correct. There is a typo in the question statement. It should be $$5e^x$$ and not $$5e^{-x}$$. Namely, in the last line of your attempt, there's indeed no negative sign for the lone $$x$$ on the right hand side.
For a plot, see Wolfram Alpha. You can see what happens when you try to change the correct 5Exp[x] to the WRONG 5Exp[-x].
Basically, you can check that $$F(x) \to 0$$ when $$x \to -\infty$$ and $$F(x) \to 1$$ when $$x \to \infty$$. Here to make this even clearer, the correct CDF is $$F(x) = 1-\frac1{(1+5 e^{x})^{1/5}}$$ as $$x \to \infty$$ the denominator blows up and makes the whole fraction vanish, resulting in the leading $$1$$ (probability one) as needed.