Inverse of $\ln(e^x-3)$

by user472288   Last Updated November 13, 2018 20:20 PM

so the whole concept about inverses is a little foggy.

Say you have function $f(x)=\ln(e^x-3)$ and you want to know the inverse function, then:

$$\ln(e^x-3) = y$$ $$e^x-3 = e^y$$ $$e^x = e^y + 3$$ $$x=\ln(e^y+3)$$ $$ f^{-1}(x)=\ln(e^x+3)$$

So my confusion comes here, the range of the original function is (-inf. to + inf), so that must be the domain of the inverse function, correct? So if we look at the domain of the inverse function by setting $e^x + 3$ to $0$, it's $\ln(-3) = x$ which is undefined and argh... what do we even have to check and define in cases of inverse functions and how should we make sure its correct... would love some clarity...

Thanks in advance.



Answers 2


The function $\log$ is increasing and $\exp$ is nonnegative, hence $\log(e^x+3) > \log(3) > 0.$

Also, your original function is defined only when $e^x-3 > 0$ or $x > \log 3.$

Will M.
Will M.
November 13, 2018 19:33 PM

Your right that the range of $f$ is from negative infinity to positive infinity. But $f^{-1}$ is also defined from negative to positive infinity, there is no problem in that since $e^x+3>0$ for all $x$

Fareed AF
Fareed AF
November 13, 2018 20:14 PM

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