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I have the following equation:

$$x = 1 - (1-y)^t$$

I would like to solve for $y$ in terms of $x$ and $t$. I tried WolframAlpha, but it did not generate a solution. Before you ask, yes this is a real problem, no this is not homework.

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    $\begingroup$ $(1-y)^t=(1-x)$. So $1-y=(1-x)^{1/t}$. $\endgroup$ Aug 14, 2017 at 21:56
  • $\begingroup$ Subtract both sides from $1$, then take logarithms of both sides. Whoops, that solves for $t$, never mind. $\endgroup$
    – MPW
    Aug 14, 2017 at 21:56
  • $\begingroup$ Logarithms are overkill, @MPW, though it works. $\endgroup$ Aug 14, 2017 at 21:57
  • $\begingroup$ @ThomasAndrews : Yes, I thought he was after the exponent. My bad. $\endgroup$
    – MPW
    Aug 14, 2017 at 21:57
  • $\begingroup$ @ThomasAndrews this is the answer, thanks. My mind has forgotten how to get rid of the superscript and account for it on the other side. $\endgroup$
    – user234105
    Aug 14, 2017 at 22:00

1 Answer 1

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$$x = 1 - (1-y)^t$$ $$(1-y)^t=1-x$$ Notice that the following step does not always hold, for example if $t=2$, and $x>1$, we end up with no solution for $y$. But suppose every value is proper, we then have: $$\bigg( (1-y)^t \bigg)^{1/t}=(1-x)^{1/t}$$ $$1-y=(1-x)^{1/t}$$ $$-y=(1-x)^{1/t} -1$$ $$y=1-(1-x)^{1/t}$$

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  • $\begingroup$ No problem! My pleasure. $\endgroup$
    – amWhy
    Aug 14, 2017 at 22:24
  • $\begingroup$ The OP states that Wolfram Alpha didn't produce the solution, but I found that Mathematica gave the correct solution instantly. $\endgroup$ Aug 14, 2017 at 23:13
  • $\begingroup$ @DavidG.Stork Sounds like Mathematica is better than wolfram alpha - but could also be the case that OP did not enter the query the way worlfram alpha understands it. $\endgroup$
    – Jay Zha
    Aug 14, 2017 at 23:20
  • $\begingroup$ @YujieZha: Yep. When I entered the query properly in WolframAlpha, the solution arose immediately. $\endgroup$ Aug 14, 2017 at 23:43
  • $\begingroup$ @DavidG.Stork That's nice, thanks for trying! $\endgroup$
    – Jay Zha
    Aug 14, 2017 at 23:44

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