Precalculus Examples

$y = 3^{x} + 2$

Step 1

Interchange the variables.

$x = 3^{y} + 2$

Step 2

Solve for $y$ .

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Rewrite the equation as $3^{y} + 2 = x$ .

$3^{y} + 2 = x$

Subtract $2$ from both sides of the equation.

$3^{y} = x - 2$

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

$\ln (3^{y}) = \ln (x - 2)$

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

$y \ln (3) = \ln (x - 2)$

Divide each term in $y \ln (3) = \ln (x - 2)$ by $\ln (3)$ and simplify.

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Divide each term in $y \ln (3) = \ln (x - 2)$ by $\ln (3)$ .

$\frac{y \ln (3)}{\ln (3)} = \frac{\ln (x - 2)}{\ln (3)}$

Simplify the left side.

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Cancel the common factor of $\ln (3)$ .

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Cancel the common factor.

$\frac{y \ln (3)}{\ln (3)} = \frac{\ln (x - 2)}{\ln (3)}$

Divide $y$ by $1$ .

$y = \frac{\ln (x - 2)}{\ln (3)}$

Step 3

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

$f^{- 1} (x) = \frac{\ln (x - 2)}{\ln (3)}$

Step 4

Verify if $f^{- 1} (x) = \frac{\ln (x - 2)}{\ln (3)}$ is the inverse of $f (x) = 3^{x} + 2$ .

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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 $f^{- 1} (3^{x} + 2)$ by substituting in the value of $f$ into $f^{- 1}$ .

$f^{- 1} (3^{x} + 2) = \frac{\ln ((3^{x} + 2) - 2)}{\ln (3)}$

Combine the opposite terms in $3^{x} + 2 - 2$ .

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Subtract $2$ from $2$ .

$f^{- 1} (3^{x} + 2) = \frac{\ln (3^{x} + 0)}{\ln (3)}$

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

$f^{- 1} (3^{x} + 2) = \frac{\ln (3^{x})}{\ln (3)}$

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

$f^{- 1} (3^{x} + 2) = \frac{x \ln (3)}{\ln (3)}$

Cancel the common factor of $\ln (3)$ .

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Cancel the common factor.

$f^{- 1} (3^{x} + 2) = \frac{x \ln (3)}{\ln (3)}$

Divide $x$ by $1$ .

$f^{- 1} (3^{x} + 2) = x$

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

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Set up the composite result function.

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

Evaluate $f (\frac{\ln (x - 2)}{\ln (3)})$ by substituting in the value of $f^{- 1}$ into $f$ .

$f (\frac{\ln (x - 2)}{\ln (3)}) = 3^{\frac{\ln (x - 2)}{\ln (3)}} + 2$

Simplify each term.

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

$f (\frac{\ln (x - 2)}{\ln (3)}) = 3^{\log_{3} (x - 2)} + 2$

Exponentiation and log are inverse functions.

$f (\frac{\ln (x - 2)}{\ln (3)}) = x - 2 + 2$

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

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Add $- 2$ and $2$ .

$f (\frac{\ln (x - 2)}{\ln (3)}) = x + 0$

Add $x$ and $0$ .

$f (\frac{\ln (x - 2)}{\ln (3)}) = x$

Since $f^{- 1} (f (x)) = x$ and $f (f^{- 1} (x)) = x$ , then $f^{- 1} (x) = \frac{\ln (x - 2)}{\ln (3)}$ is the inverse of $f (x) = 3^{x} + 2$ .

$f^{- 1} (x) = \frac{\ln (x - 2)}{\ln (3)}$