Find the Inverse y=e^(x+3)-4

$y = e^{x + 3} - 4$

Step 1

Interchange the variables.

$x = e^{y + 3} - 4$

Step 2

Solve for $y$ .

Rewrite the equation as $e^{y + 3} - 4 = x$ .

$e^{y + 3} - 4 = x$

Add $4$ to both sides of the equation.

$e^{y + 3} = x + 4$

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

$\ln (e^{y + 3}) = \ln (x + 4)$

Expand the left side.

Expand $\ln (e^{y + 3})$ by moving $y + 3$ outside the logarithm.

$(y + 3) \ln (e) = \ln (x + 4)$

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

$(y + 3) \cdot 1 = \ln (x + 4)$

Multiply $y + 3$ by $1$ .

$y + 3 = \ln (x + 4)$

Subtract $3$ from both sides of the equation.

$y = \ln (x + 4) - 3$

Step 3

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

$f^{- 1} (x) = \ln (x + 4) - 3$

Step 4

Verify if $f^{- 1} (x) = \ln (x + 4) - 3$ is the inverse of $f (x) = e^{x + 3} - 4$ .

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 $f^{- 1} (e^{x + 3} - 4)$ by substituting in the value of $f$ into $f^{- 1}$ .

$f^{- 1} (e^{x + 3} - 4) = \ln ((e^{x + 3} - 4) + 4) - 3$

Combine the opposite terms in $\ln ((e^{x + 3} - 4) + 4) - 3$ .

Add $- 4$ and $4$ .

$f^{- 1} (e^{x + 3} - 4) = \ln (e^{x + 3} + 0) - 3$

Add $e^{x + 3}$ and $0$ .

$f^{- 1} (e^{x + 3} - 4) = \ln (e^{x + 3}) - 3$

Simplify each term.

Use logarithm rules to move $x + 3$ out of the exponent.

$f^{- 1} (e^{x + 3} - 4) = (x + 3) \ln (e) - 3$

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

$f^{- 1} (e^{x + 3} - 4) = (x + 3) \cdot 1 - 3$

Multiply $x + 3$ by $1$ .

$f^{- 1} (e^{x + 3} - 4) = x + 3 - 3$

Combine the opposite terms in $x + 3 - 3$ .

Subtract $3$ from $3$ .

$f^{- 1} (e^{x + 3} - 4) = x + 0$

Add $x$ and $0$ .

$f^{- 1} (e^{x + 3} - 4) = x$

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

Set up the composite result function.

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

Evaluate $f (\ln (x + 4) - 3)$ by substituting in the value of $f^{- 1}$ into $f$ .

$f (\ln (x + 4) - 3) = e^{(\ln (x + 4) - 3) + 3} - 4$

Combine the opposite terms in $e^{(\ln (x + 4) - 3) + 3} - 4$ .

Add $- 3$ and $3$ .

$f (\ln (x + 4) - 3) = e^{\ln (x + 4) + 0} - 4$

Add $\ln (x + 4)$ and $0$ .

$f (\ln (x + 4) - 3) = e^{\ln (x + 4)} - 4$

Exponentiation and log are inverse functions.

$f (\ln (x + 4) - 3) = x + 4 - 4$

Combine the opposite terms in $x + 4 - 4$ .

Subtract $4$ from $4$ .

$f (\ln (x + 4) - 3) = x + 0$

Add $x$ and $0$ .

$f (\ln (x + 4) - 3) = x$

Since $f^{- 1} (f (x)) = x$ and $f (f^{- 1} (x)) = x$ , then $f^{- 1} (x) = \ln (x + 4) - 3$ is the inverse of $f (x) = e^{x + 3} - 4$ .

$f^{- 1} (x) = \ln (x + 4) - 3$

Find the Inverse y=e^(x+3)-4

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