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I am trying to find out the inverse function of $y=\left(\frac{1}{8}\right)^{1-x}$.

Here's a picture of I've got so far: $$x=(1/8)^{1-y}$$ $$e^x=e^{(1-y)\ln1/8}=(1-y)\ln\frac18$$ $$\ln8e^x=1-y$$ $$y=1-\ln8e^x$$

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    $\begingroup$ The inverse of a function depends on which range and domain you consider; please state these in order to determine the inverse function $\endgroup$
    – Victor
    Sep 29, 2022 at 18:08
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    $\begingroup$ In your solution .jpg, there are serious mistakes. $\endgroup$
    – Bob Dobbs
    Sep 29, 2022 at 18:25
  • $\begingroup$ What are they ? That would really help me $\endgroup$
    – Adam Cora
    Sep 29, 2022 at 18:26
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    $\begingroup$ @AdamCora that is in the second step. it should've been $e^{\ln x}=e^{(1-y)\ln \frac{1}{8}}$ $\endgroup$ Sep 29, 2022 at 19:48

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Switching variables and taking the logarithm base $\frac{1}{8}$ of both sides: $$1-y=\log_{\frac{1}{8}}(x)$$ Then solving for $y$: $$y=1-\log_{\frac{1}{8}}(x)$$ Logarithms are the inverse of exponentials. So for example: $$a=b^x \implies \log_ba=x$$

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Third line in your .jpg file is mistake. It must be: $$x=e^{\ln\left(\frac{1}{8}\right)(1-y)}$$ Then take "$\ln$" of both sides to get $$\ln x = \ln\left(\frac{1}{8}\right)(1-y).$$ Then, since $\ln(\frac{1}{8})=-\ln8$, we have $$\ln x = \ln8(y-1)$$ or $$\frac{\ln x}{\ln 8}=y-1.$$ Hence, $$y=\frac{1}{\ln 8}\ln x+1$$ is the solution. You may write it in base 8 logarithm as $y=\log_8 x +1$.

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  • $\begingroup$ Shouldn't the 8 be at the top since we're dividing by 8 and not multiplying by 8 ? $\endgroup$
    – Adam Cora
    Sep 29, 2022 at 19:52
  • $\begingroup$ There is no 8. You mean $\ln 8$? $\endgroup$
    – Bob Dobbs
    Sep 29, 2022 at 20:15
  • $\begingroup$ Yeah that's what I meant $\endgroup$
    – Adam Cora
    Sep 29, 2022 at 20:25
  • $\begingroup$ $\frac{\ln x}{\ln a} = \log_{a} x$ @AdamCora $\endgroup$
    – Bob Dobbs
    Sep 29, 2022 at 21:13
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As a general rule, when dealing with variables in the exponentials that you want to get down, use the log function.
$$ log_{1/8}y =1-x => x= 1- log_{1/8}y $$

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