Find the Inverse y=e^x

$y = e^{x}$

Interchange the variables.

$x = e^{y}$

Solve for $y$ .

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

$e^{y} = x$

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

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

Expand the left side.

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

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

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

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

Multiply $y$ by $1$ .

$y = \ln (x)$

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

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

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

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

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

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

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

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

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

Multiply $x$ by $1$ .

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

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

Set up the composite result function.

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

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

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

Exponentiation and log are inverse functions.

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

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

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