Find the Inverse y=(4^x+8)/4

$y = \frac{4^{x} + 8}{4}$

Step 1

Interchange the variables.

$x = \frac{4^{y} + 8}{4}$

Step 2

Solve for $y$ .

Rewrite the equation as $\frac{4^{y} + 8}{4} = x$ .

$\frac{4^{y} + 8}{4} = x$

Multiply both sides of the equation by $4$ .

$4 \frac{4^{y} + 8}{4} = 4 x$

Simplify the left side.

Cancel the common factor of $4$ .

Cancel the common factor.

$4 \frac{4^{y} + 8}{4} = 4 x$

Rewrite the expression.

$4^{y} + 8 = 4 x$

Subtract $8$ from both sides of the equation.

$4^{y} = 4 x - 8$

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

$\ln (4^{y}) = \ln (4 x - 8)$

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

$y \ln (4) = \ln (4 x - 8)$

Divide each term in $y \ln (4) = \ln (4 x - 8)$ by $\ln (4)$ and simplify.

Divide each term in $y \ln (4) = \ln (4 x - 8)$ by $\ln (4)$ .

$\frac{y \ln (4)}{\ln (4)} = \frac{\ln (4 x - 8)}{\ln (4)}$

Simplify the left side.

Cancel the common factor of $\ln (4)$ .

Cancel the common factor.

$\frac{y \ln (4)}{\ln (4)} = \frac{\ln (4 x - 8)}{\ln (4)}$

Divide $y$ by $1$ .

$y = \frac{\ln (4 x - 8)}{\ln (4)}$

Step 3

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

$f^{- 1} (x) = \frac{\ln (4 x - 8)}{\ln (4)}$

Step 4

Verify if $f^{- 1} (x) = \frac{\ln (4 x - 8)}{\ln (4)}$ is the inverse of $f (x) = \frac{4^{x} + 8}{4}$ .

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 $f^{- 1} (\frac{4^{x} + 8}{4})$ by substituting in the value of $f$ into $f^{- 1}$ .

$f^{- 1} (\frac{4^{x} + 8}{4}) = \frac{\ln (4 (\frac{4^{x} + 8}{4}) - 8)}{\ln (4)}$

Simplify the numerator.

Cancel the common factor of $4$ .

Cancel the common factor.

$f^{- 1} (\frac{4^{x} + 8}{4}) = \frac{\ln (4 (\frac{4^{x} + 8}{4}) - 8)}{\ln (4)}$

Rewrite the expression.

$f^{- 1} (\frac{4^{x} + 8}{4}) = \frac{\ln (4^{x} + 8 - 8)}{\ln (4)}$

Combine the opposite terms in $4^{x} + 8 - 8$ .

Subtract $8$ from $8$ .

$f^{- 1} (\frac{4^{x} + 8}{4}) = \frac{\ln (4^{x} + 0)}{\ln (4)}$

Add $4^{x}$ and $0$ .

$f^{- 1} (\frac{4^{x} + 8}{4}) = \frac{\ln (4^{x})}{\ln (4)}$

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

$f^{- 1} (\frac{4^{x} + 8}{4}) = \frac{x \ln (4)}{\ln (4)}$

Cancel the common factor of $\ln (4)$ .

Cancel the common factor.

$f^{- 1} (\frac{4^{x} + 8}{4}) = \frac{x \ln (4)}{\ln (4)}$

Divide $x$ by $1$ .

$f^{- 1} (\frac{4^{x} + 8}{4}) = x$

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

Set up the composite result function.

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

Evaluate $f (\frac{\ln (4 x - 8)}{\ln (4)})$ by substituting in the value of $f^{- 1}$ into $f$ .

$f (\frac{\ln (4 x - 8)}{\ln (4)}) = \frac{4^{\frac{\ln (4 x - 8)}{\ln (4)}} + 8}{4}$

Simplify the numerator.

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

$f (\frac{\ln (4 x - 8)}{\ln (4)}) = \frac{4^{\log_{4} (4 x - 8)} + 8}{4}$

Exponentiation and log are inverse functions.

$f (\frac{\ln (4 x - 8)}{\ln (4)}) = \frac{4 x - 8 + 8}{4}$

Add $- 8$ and $8$ .

$f (\frac{\ln (4 x - 8)}{\ln (4)}) = \frac{4 x + 0}{4}$

Add $4 x$ and $0$ .

$f (\frac{\ln (4 x - 8)}{\ln (4)}) = \frac{4 x}{4}$

Cancel the common factor of $4$ .

Cancel the common factor.

$f (\frac{\ln (4 x - 8)}{\ln (4)}) = \frac{4 x}{4}$

Divide $x$ by $1$ .

$f (\frac{\ln (4 x - 8)}{\ln (4)}) = x$

Since $f^{- 1} (f (x)) = x$ and $f (f^{- 1} (x)) = x$ , then $f^{- 1} (x) = \frac{\ln (4 x - 8)}{\ln (4)}$ is the inverse of $f (x) = \frac{4^{x} + 8}{4}$ .

$f^{- 1} (x) = \frac{\ln (4 x - 8)}{\ln (4)}$

Find the Inverse y=(4^x+8)/4

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