# Challenging: The inverse function of $y=\left(\frac{1}{8}\right)^{1-x}$

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: $$x=(1/8)^{1-y}$$ $$e^x=e^{(1-y)\ln1/8}=(1-y)\ln\frac18$$ $$\ln8e^x=1-y$$ $$y=1-\ln8e^x$$

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

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: $$a=b^x \implies \log_ba=x$$
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 $$y=\log_8 x +1$$.
• There is no 8. You mean $\ln 8$? Sep 29, 2022 at 20:15
• $\frac{\ln x}{\ln a} = \log_{a} x$ @AdamCora Sep 29, 2022 at 21:13
$$log_{1/8}y =1-x => x= 1- log_{1/8}y$$