Violet Woodward

2022-07-18

Inverse of the function $-\mathrm{log}\left(1-\left[1-{e}^{-{x}^{\alpha }}{\right]}^{\beta }\right)$
I have a function as follows, I would like to get the inverse of this function. What is the inverse of $f\left(x\right)$?
$y=f\left(x\right)=-\mathrm{log}\left(1-\left[1-{e}^{-{x}^{\alpha }}{\right]}^{\beta }\right)$
${f}^{-1}\left(x\right)=\left(-\mathrm{log}\left(1-\left(1-{\mathrm{exp}}^{-x}{\right)}^{1/\beta }\right){\right)}^{1/\alpha }$

decoratesuw

Expert

Since you have already done the first step ..
$y=f\left(x\right)=-\mathrm{log}\left(1-\left[1-\mathrm{exp}\left(-{x}^{\alpha }\right){\right]}^{\beta }\right)$
$\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}-y=\mathrm{log}\left(1-\left[1-{e}^{-{x}^{\alpha }}{\right]}^{\beta }\right)$
$\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}{e}^{-y}=1-\left[1-{e}^{-{x}^{\alpha }}{\right]}^{\beta }\right)$
$\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}-{e}^{-y}+1=\left[1-{e}^{-{x}^{\alpha }}{\right]}^{\beta }$
$\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}\left[1-{e}^{-y}{\right]}^{\frac{1}{\beta }}-1=-{e}^{-{x}^{\alpha }}$
which turns out to be
${e}^{-{x}^{\alpha }}=1-\left[1-{e}^{-y}{\right]}^{\frac{1}{\beta }}$
. Taking $\mathrm{log}$ on both sides we get
$-{x}^{\alpha }=\mathrm{log}\left(1-\left[1-{e}^{-y}{\right]}^{\frac{1}{\beta }}\right)\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}log\frac{1}{\left(1-\left[1-{e}^{-y}{\right]}^{\frac{1}{\beta }}\right)}={x}^{\alpha }$
Hence
${f}^{-1}\left(x\right)={\left(log\frac{1}{\left(1-\left[1-{e}^{-x}{\right]}^{\frac{1}{\beta }}\right)}\right)}^{\frac{1}{\alpha }}$

Raegan Bray

Expert

Multiply both sides by $-1$, exponentiate both sides, subtract 1 from both sides, multiply both sides by $-1$, raise both sides to the $\frac{1}{\beta }$ power, subtract 1 from both sides, multiply both sides by $-1$, take natural logarithm of both sides, multiply both sides by $-1$, raise both sides to $\frac{1}{\alpha }$ power.
This solves for $x$ in terms of $y$, which gives you the inverse function in terms of $y$
(Note that you will need the $\frac{1}{\beta }$ and $\frac{1}{\alpha }$ powers to be defined)

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