# What is the inverse of y= e^(x-1)-1 ?

Dec 7, 2015

${f}^{- 1} \left(x\right) = \ln \left(x + 1\right) + 1$

#### Explanation:

To compute the inverse, you need to follow the following steps:

1) swap $y$ and $x$ in your equation:

$x = {e}^{y - 1} - 1$

2) solve the equation for $y$:

... add $1$ on both sides of the equation...

$x + 1 = {e}^{y - 1}$

... remember that $\ln x$ is the inverse function for ${e}^{x}$ which means that both $\ln \left({e}^{x}\right) = x$ and ${e}^{\ln x} = x$ hold.
This means that you can apply $\ln \left(\right)$ on both sides of the equation to "get rid" of the exponential function:

$\ln \left(x + 1\right) = \ln \left({e}^{y - 1}\right)$

$\ln \left(x + 1\right) = y - 1$

... add $1$ on both sides of the equation again...

$\ln \left(x + 1\right) + 1 = y$

3) Now, just replace $y$ with ${f}^{- 1} \left(x\right)$ and you have the result!

So, for

$f \left(x\right) = {e}^{x - 1} - 1$,

the inverse function is

${f}^{- 1} \left(x\right) = \ln \left(x + 1\right) + 1$

Hope that this helped!