Find the Inverse f(x)=e^x-1

$f (x) = e^{x} - 1$

Write $f (x) = e^{x} - 1$ as an equation.

$y = e^{x} - 1$

Interchange the variables.

$x = e^{y} - 1$

Solve for $y$ .

Rewrite the equation as $e^{y} - 1 = x$ .

$e^{y} - 1 = x$

Add $1$ to both sides of the equation.

$e^{y} = x + 1$

Take the natural logarithm of both sides of the equation to remove the variable from the exponent.

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

Expand the left side.

Expand $\ln (e^{y})$ by moving $y$ outside the logarithm.

$y \ln (e) = \ln (x + 1)$

The natural logarithm of $e$ is $1$ .

$y \cdot 1 = \ln (x + 1)$

Multiply $y$ by $1$ .

$y = \ln (x + 1)$

Replace $y$ with $f^{- 1} (x)$ to show the final answer.

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

Verify if $f^{- 1} (x) = \ln (x + 1)$ is the inverse of $f (x) = e^{x} - 1$ .

To verify the inverse, check if $f^{- 1} (f (x)) = x$ and $f (f^{- 1} (x)) = x$ .

Evaluate $f^{- 1} (f (x))$ .

Set up the composite result function.

$f^{- 1} (f (x))$

Evaluate $f^{- 1} (e^{x} - 1)$ by substituting in the value of $f$ into $f^{- 1}$ .

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

Combine the opposite terms in $(e^{x} - 1) + 1$ .

Add $- 1$ and $1$ .

$f^{- 1} (e^{x} - 1) = \ln (e^{x} + 0)$

Add $e^{x}$ and $0$ .

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

Use logarithm rules to move $x$ out of the exponent.

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

The natural logarithm of $e$ is $1$ .

$f^{- 1} (e^{x} - 1) = x \cdot 1$

Multiply $x$ by $1$ .

$f^{- 1} (e^{x} - 1) = x$

Evaluate $f (f^{- 1} (x))$ .

Set up the composite result function.

$f (f^{- 1} (x))$

Evaluate $f (\ln (x + 1))$ by substituting in the value of $f^{- 1}$ into $f$ .

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

Exponentiation and log are inverse functions.

$f (\ln (x + 1)) = x + 1 - 1$

Combine the opposite terms in $x + 1 - 1$ .

Subtract $1$ from $1$ .

$f (\ln (x + 1)) = x + 0$

Add $x$ and $0$ .

$f (\ln (x + 1)) = x$

Since $f^{- 1} (f (x)) = x$ and $f (f^{- 1} (x)) = x$ , then $f^{- 1} (x) = \ln (x + 1)$ is the inverse of $f (x) = e^{x} - 1$ .

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

Find the Inverse f(x)=e^x-1

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