Violet Woodward

Answered

2022-07-18

Inverse of the function $-\mathrm{log}(1-[1-{e}^{-{x}^{\alpha}}{]}^{\beta})$

I have a function as follows, I would like to get the inverse of this function. What is the inverse of $f(x)$?

$y=f(x)=-\mathrm{log}(1-[1-{e}^{-{x}^{\alpha}}{]}^{\beta})$

Is my answer correct?

${f}^{-1}(x)=(-\mathrm{log}(1-(1-{\mathrm{exp}}^{-x}{)}^{1/\beta}){)}^{1/\alpha}$

I have a function as follows, I would like to get the inverse of this function. What is the inverse of $f(x)$?

$y=f(x)=-\mathrm{log}(1-[1-{e}^{-{x}^{\alpha}}{]}^{\beta})$

Is my answer correct?

${f}^{-1}(x)=(-\mathrm{log}(1-(1-{\mathrm{exp}}^{-x}{)}^{1/\beta}){)}^{1/\alpha}$

Answer & Explanation

decoratesuw

Expert

2022-07-19Added 11 answers

Since you have already done the first step ..

$y=f(x)=-\mathrm{log}(1-[1-\mathrm{exp}(-{x}^{\alpha}){]}^{\beta})$

$\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}-y=\mathrm{log}(1-[1-{e}^{-{x}^{\alpha}}{]}^{\beta})$

$\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}{e}^{-y}=1-[1-{e}^{-{x}^{\alpha}}{]}^{\beta})$

$\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}-{e}^{-y}+1=[1-{e}^{-{x}^{\alpha}}{]}^{\beta}$

$\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}[1-{e}^{-y}{]}^{\frac{1}{\beta}}-1=-{e}^{-{x}^{\alpha}}$

which turns out to be

${e}^{-{x}^{\alpha}}=1-[1-{e}^{-y}{]}^{\frac{1}{\beta}}$

. Taking $\mathrm{log}$ on both sides we get

$-{x}^{\alpha}=\mathrm{log}(1-[1-{e}^{-y}{]}^{\frac{1}{\beta}})\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}log\frac{1}{(1-[1-{e}^{-y}{]}^{\frac{1}{\beta}})}={x}^{\alpha}$

Hence

${f}^{-1}(x)={\left(log\frac{1}{(1-[1-{e}^{-x}{]}^{\frac{1}{\beta}})}\right)}^{\frac{1}{\alpha}}$

$y=f(x)=-\mathrm{log}(1-[1-\mathrm{exp}(-{x}^{\alpha}){]}^{\beta})$

$\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}-y=\mathrm{log}(1-[1-{e}^{-{x}^{\alpha}}{]}^{\beta})$

$\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}{e}^{-y}=1-[1-{e}^{-{x}^{\alpha}}{]}^{\beta})$

$\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}-{e}^{-y}+1=[1-{e}^{-{x}^{\alpha}}{]}^{\beta}$

$\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}[1-{e}^{-y}{]}^{\frac{1}{\beta}}-1=-{e}^{-{x}^{\alpha}}$

which turns out to be

${e}^{-{x}^{\alpha}}=1-[1-{e}^{-y}{]}^{\frac{1}{\beta}}$

. Taking $\mathrm{log}$ on both sides we get

$-{x}^{\alpha}=\mathrm{log}(1-[1-{e}^{-y}{]}^{\frac{1}{\beta}})\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}log\frac{1}{(1-[1-{e}^{-y}{]}^{\frac{1}{\beta}})}={x}^{\alpha}$

Hence

${f}^{-1}(x)={\left(log\frac{1}{(1-[1-{e}^{-x}{]}^{\frac{1}{\beta}})}\right)}^{\frac{1}{\alpha}}$

Raegan Bray

Expert

2022-07-20Added 1 answers

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)

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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