Find the Inverse 5^(x+4)=y

$5^{x + 4} = y$

Rewrite the equation as $y = 5^{x + 4}$ .

$y = 5^{x + 4}$

Interchange the variables.

$x = 5^{y + 4}$

Solve for $y$ .

Rewrite the equation as $5^{y + 4} = x$ .

$5^{y + 4} = x$

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

$\ln (5^{y + 4}) = \ln (x)$

Expand $\ln (5^{y + 4})$ by moving $y + 4$ outside the logarithm.

$(y + 4) \ln (5) = \ln (x)$

Simplify the left side.

Apply the distributive property.

$y \ln (5) + 4 \ln (5) = \ln (x)$

Move all the terms containing a logarithm to the left side of the equation.

$y \ln (5) + 4 \ln (5) - \ln (x) = 0$

Move all terms not containing $y$ to the right side of the equation.

Subtract $4 \ln (5)$ from both sides of the equation.

$y \ln (5) - \ln (x) = - 4 \ln (5)$

Add $\ln (x)$ to both sides of the equation.

$y \ln (5) = - 4 \ln (5) + \ln (x)$

Divide each term in $y \ln (5) = - 4 \ln (5) + \ln (x)$ by $\ln (5)$ and simplify.

Divide each term in $y \ln (5) = - 4 \ln (5) + \ln (x)$ by $\ln (5)$ .

$\frac{y \ln (5)}{\ln (5)} = \frac{- 4 \ln (5)}{\ln (5)} + \frac{\ln (x)}{\ln (5)}$

Simplify the left side.

Cancel the common factor of $\ln (5)$ .

Cancel the common factor.

$\frac{y \ln (5)}{\ln (5)} = \frac{- 4 \ln (5)}{\ln (5)} + \frac{\ln (x)}{\ln (5)}$

Divide $y$ by $1$ .

$y = \frac{- 4 \ln (5)}{\ln (5)} + \frac{\ln (x)}{\ln (5)}$

Simplify the right side.

Cancel the common factor of $\ln (5)$ .

Cancel the common factor.

$y = \frac{- 4 \ln (5)}{\ln (5)} + \frac{\ln (x)}{\ln (5)}$

Divide $- 4$ by $1$ .

$y = - 4 + \frac{\ln (x)}{\ln (5)}$

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

$f^{- 1} (x) = - 4 + \frac{\ln (x)}{\ln (5)}$

Verify if $f^{- 1} (x) = - 4 + \frac{\ln (x)}{\ln (5)}$ is the inverse of $f (x) = 5^{x + 4}$ .

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 $f^{- 1} \cdot 5^{x + 4}$ by substituting in the value of $f$ into $f^{- 1}$ .

$f^{- 1} \cdot 5^{x + 4} = - 4 + \frac{\ln (5^{x + 4})}{\ln (5)}$

Simplify each term.

Expand $\ln (5^{x + 4})$ by moving $x + 4$ outside the logarithm.

$f^{- 1} \cdot 5^{x + 4} = - 4 + \frac{(x + 4) \ln (5)}{\ln (5)}$

Cancel the common factor of $\ln (5)$ .

Cancel the common factor.

$f^{- 1} \cdot 5^{x + 4} = - 4 + \frac{(x + 4) \ln (5)}{\ln (5)}$

Divide $x + 4$ by $1$ .

$f^{- 1} \cdot 5^{x + 4} = - 4 + x + 4$

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

Add $- 4$ and $4$ .

$f^{- 1} \cdot 5^{x + 4} = x + 0$

Add $x$ and $0$ .

$f^{- 1} \cdot 5^{x + 4} = x$

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

Set up the composite result function.

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

Evaluate $f (- 4 + \frac{\ln (x)}{\ln (5)})$ by substituting in the value of $f^{- 1}$ into $f$ .

$f (- 4 + \frac{\ln (x)}{\ln (5)}) = 5^{(- 4 + \frac{\ln (x)}{\ln (5)}) + 4}$

Combine the opposite terms in $(- 4 + \frac{\ln (x)}{\ln (5)}) + 4$ .

Add $- 4$ and $4$ .

$f (- 4 + \frac{\ln (x)}{\ln (5)}) = 5^{0 + \frac{\ln (x)}{\ln (5)}}$

Add $0$ and $\frac{\ln (x)}{\ln (5)}$ .

$f (- 4 + \frac{\ln (x)}{\ln (5)}) = 5^{\frac{\ln (x)}{\ln (5)}}$

Use the change of base rule $\frac{\log_{b} (x)}{\log_{b} (a)} = \log_{a} (x)$ .

$f (- 4 + \frac{\ln (x)}{\ln (5)}) = 5^{\log_{5} (x)}$

Exponentiation and log are inverse functions.

$f (- 4 + \frac{\ln (x)}{\ln (5)}) = x$

Since $f^{- 1} (f (x)) = x$ and $f (f^{- 1} (x)) = x$ , then $f^{- 1} (x) = - 4 + \frac{\ln (x)}{\ln (5)}$ is the inverse of $f (x) = 5^{x + 4}$ .

$f^{- 1} (x) = - 4 + \frac{\ln (x)}{\ln (5)}$

Find the Inverse 5^(x+4)=y

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