Precalculus Examples

$y = e^{x + 2}$

Interchange the variables.

$x = e^{y + 2}$

Solve for $y$ .

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Rewrite the equation as $e^{y + 2} = x$ .

$e^{y + 2} = x$

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

$\ln (e^{y + 2}) = \ln (x)$

Expand the left side.

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Expand $\ln (e^{y + 2})$ by moving $y + 2$ outside the logarithm.

$(y + 2) \ln (e) = \ln (x)$

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

$(y + 2) \cdot 1 = \ln (x)$

Multiply $y + 2$ by $1$ .

$y + 2 = \ln (x)$

Subtract $2$ from both sides of the equation.

$y = \ln (x) - 2$

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

$f^{- 1} (x) = \ln (x) - 2$

Verify if $f^{- 1} (x) = \ln (x) - 2$ is the inverse of $f (x) = e^{x + 2}$ .

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

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

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Set up the composite result function.

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

Evaluate $f^{- 1} (e^{x + 2})$ by substituting in the value of $f$ into $f^{- 1}$ .

$f^{- 1} (e^{x + 2}) = \ln (e^{x + 2}) - 2$

Simplify each term.

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Use logarithm rules to move $x + 2$ out of the exponent.

$f^{- 1} (e^{x + 2}) = (x + 2) \ln (e) - 2$

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

$f^{- 1} (e^{x + 2}) = (x + 2) \cdot 1 - 2$

Multiply $x + 2$ by $1$ .

$f^{- 1} (e^{x + 2}) = x + 2 - 2$

Combine the opposite terms in $x + 2 - 2$ .

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Subtract $2$ from $2$ .

$f^{- 1} (e^{x + 2}) = x + 0$

Add $x$ and $0$ .

$f^{- 1} (e^{x + 2}) = x$

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

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Set up the composite result function.

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

Evaluate $f (\ln (x) - 2)$ by substituting in the value of $f^{- 1}$ into $f$ .

$f (\ln (x) - 2) = e^{(\ln (x) - 2) + 2}$

Combine the opposite terms in $(\ln (x) - 2) + 2$ .

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Add $- 2$ and $2$ .

$f (\ln (x) - 2) = e^{\ln (x) + 0}$

Add $\ln (x)$ and $0$ .

$f (\ln (x) - 2) = e^{\ln (x)}$

Exponentiation and log are inverse functions.

$f (\ln (x) - 2) = x$

Since $f^{- 1} (f (x)) = x$ and $f (f^{- 1} (x)) = x$ , then $f^{- 1} (x) = \ln (x) - 2$ is the inverse of $f (x) = e^{x + 2}$ .

$f^{- 1} (x) = \ln (x) - 2$