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How do I solve this for $y$?

$$u= 1 - \exp\left\{-\left(\frac{y-\theta}{\alpha}\right)^\gamma\right\}.$$

If I take the $\log$ I end up with

$$\log(1-u) = -\left(\frac{y-\theta}{\alpha}\right)^\gamma.$$ I'm unclear on how to proceed further to solve the equation for $y$. I apologize for not using the correct math symbols.

Michael Hardy
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user3536311
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  • Subtract $1$ from both sides and then multiply both sides by $-1$ to get an equation of the form $1 - u = e^{\mu}$, which then suggests taking a logarithm. – Travis Willse Jan 30 '16 at 22:51

1 Answers1

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I assume $\gamma > 0$. You have $$ \log(1-u) = -\left(\frac{y-\theta}{\alpha}\right)^\gamma $$ Multiply by $-1$ (so that both sides are positive) and take the $1/\gamma$-th power on both sides to get $$ (-\log(1-u))^{1/\gamma} = \frac{y-\theta}{\alpha} $$ Now, rearranging (multiply both sides by $\alpha$, then add $\theta$), you get $$ \alpha(-\log(1-u))^{1/\gamma} + \theta = y $$ or equivalently $$ y = \alpha\left(\log\frac{1}{1-u}\right)^{1/\gamma} + \theta $$

Note: all steps are fine and can be rewritten as equivalences as the function $x\geq 0\mapsto x^{1/\gamma}$ is a bijection.

Clement C.
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