I try to calculate the inverse of the following exponential function (it's a CDF): $$ f(x) = { 1 \over 3} e^{-x/2} + { 2 \over 3} e^{-x/4} $$
I used a website to calculate it, they provided me the following: $$f^{-1}(x) = 4\cdot \ln\left({\sqrt{3x+1} +1 \over x}\right) + 4 \cdot \ln(3), $$ which seems to be correct. I now try to understand the way it was calculated but I can't figure it out.
I started with:
\begin{align} y &= { 1 \over 3} e^{-x/2} + { 2 \over 3} e^{-x/4}\\ y &= 3 \cdot ( e^{-x/2} + 2 \cdot e^{-x/4} )\\ \ln(y) &= \ln(3 \cdot ( e^{-x/2} + 2 \cdot e^{-x/4} ))\\ \ln(y) - \ln(3) &= \ln( e^{-x/2} + 2 \cdot e^{-x/4} ) \end{align}
At this point I tried $\ln(a + c) = \ln(a) + \ln(1 + { c \over a}) $ which gave me:
\begin{align} \ln(y) - \ln(3) &= -{x \over 2} + \ln\left(1 + 2 \cdot {e^{-x/4} \over e^{-x/2} }\right)\\ &= -{x \over 2} + \ln(1 + 2 \cdot {e^{x/4}}) \end{align} At this point, I'm stuck. Any help is appreciated.
Regards
Hint. Let $z=e^{-x/4}$ and solve the quadratic equation with respect to $z$. $$y = { 1 \over 3} e^{-x/2} + { 2 \over 3} e^{-x/4}={ 1 \over 3} z^2 + { 2 \over 3} z\Leftrightarrow (z+1)^2=1+3y.$$
P.S. As pointed out by Yves Daoust, starting from the hint we easily find $$x=f^{-1}(y)=-4\log\left(\sqrt{3y+1}-1\right)$$ which can be written also in the following way \begin{align} x=f^{-1}(y)&=-4\log\left(\sqrt{3y+1}-1\right)=4\log\left(\frac{1}{\sqrt{3y+1}-1}\right)\\ &=4\log\left(\frac{\sqrt{3y+1}+1}{(3y+1)-1}\right) =4\log\left(\frac{\sqrt{3y+1}+1}{y}\right)-4\log(3). \end{align} So there is a missing minus sign in your formula. Note that $f(0)=1$ and therefore $f^{-1}(1)=0$.