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I have this function: $$ f(x)=\frac{1+\ln(x)}{1-\ln(x)} $$ And i should calculate $f^{-1}(x)$

I am not really sure how to proceed but I think that the first step would be to have x alone, how do I achieve that?

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Note, that when inversing a function you should also determine the set over which the inverse function is defined.

Set $y=f(x)$ and solve for $x$: $$y=\frac{1+\ln x}{1- \ln x} \Rightarrow y(1- \ln x)=1+\ln x \Rightarrow y-1=y(\ln x)+\ln x$$ which gives $$(y-1)=(y+1)\ln x\Rightarrow \ln x=\frac{y-1}{y+1}$$ which gives by exponentiating both sides to the power $e$ $$e^{\ln x}=e^{\frac{y-1}{y+1}}$$ which reduces to $$x=e^{\frac{y-1}{y+1}}$$ Therefore $$f^{-1}(y)=e^{\frac{y-1}{y+1}}$$ Note, also that this holds for all $y \in \mathbb{R}\backslash\{-1\}$.

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  • $\begingroup$ I did the same mistake while reading the Question $\endgroup$ – lab bhattacharjee Mar 11 '14 at 11:22
  • $\begingroup$ @labbhattacharjee Thanks, probably I was looking at your answer to proofcheck mine, so I made the same mistake :). I only posted to add the domain of $f^{-1}$ and make the first steps a little slower. Otherwise, your answer had already covered the OP's question! $\endgroup$ – Jimmy R. Mar 11 '14 at 11:28
  • $\begingroup$ Could you explain the last $\Rightarrow$ at the first row? I do not get the transformation there... $\endgroup$ – theva Mar 11 '14 at 11:58
  • $\begingroup$ @theva I edited it, is it ok now? $\endgroup$ – Jimmy R. Mar 11 '14 at 12:08
  • $\begingroup$ yes it is, thanks! $\endgroup$ – theva Mar 11 '14 at 12:55
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Let $\displaystyle f(x)=\frac{1+\ln x}{1-\ln x}=\frac y1$

$\displaystyle\implies f^{-1}(y)=x$

Applying Componendo & Dividendo,

$$ \ln x=\dfrac{y-1}{y+1}$$

$\displaystyle\implies x=e^{\left(\dfrac{y-1}{y+1}\right)}$ which is $f^{-1}(y)$

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    $\begingroup$ beat me to it. Damn. +1 $\endgroup$ – Guy Mar 11 '14 at 11:17
  • $\begingroup$ @GitGud, is there any discernible mistake ? $\endgroup$ – lab bhattacharjee Mar 11 '14 at 11:18
  • $\begingroup$ @labbhattacharjee Your $f$ is the OP's $1/f$. $\endgroup$ – Git Gud Mar 11 '14 at 11:19
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    $\begingroup$ @GitGud, thanks for your observation $\endgroup$ – lab bhattacharjee Mar 11 '14 at 11:20

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