Precalculus Examples

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

Step 1

Write $f (x) = 3^{x + 1} - 2$ as an equation.

$y = 3^{x + 1} - 2$

Step 2

Interchange the variables.

$x = 3^{y + 1} - 2$

Step 3

Solve for $y$ .

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Step 3.1

Rewrite the equation as $3^{y + 1} - 2 = x$ .

$3^{y + 1} - 2 = x$

Step 3.2

Add $2$ to both sides of the equation.

$3^{y + 1} = x + 2$

Step 3.3

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

$\ln (3^{y + 1}) = \ln (x + 2)$

Step 3.4

Expand $\ln (3^{y + 1})$ by moving $y + 1$ outside the logarithm.

$(y + 1) \ln (3) = \ln (x + 2)$

Step 3.5

Simplify the left side.

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Step 3.5.1

Simplify $(y + 1) \ln (3)$ .

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Step 3.5.1.1

Apply the distributive property.

$y \ln (3) + 1 \ln (3) = \ln (x + 2)$

Step 3.5.1.2

Multiply $\ln (3)$ by $1$ .

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

Step 3.6

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

$y \ln (3) + \ln (3) - \ln (x + 2) = 0$

Step 3.7

Use the quotient property of logarithms, $\log_{b} (x) - \log_{b} (y) = \log_{b} (\frac{x}{y})$ .

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

Step 3.8

Subtract $\ln (\frac{3}{x + 2})$ from both sides of the equation.

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

Step 3.9

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

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Step 3.9.1

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

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

Step 3.9.2

Simplify the left side.

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Step 3.9.2.1

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

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Step 3.9.2.1.1

Cancel the common factor.

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

Step 3.9.2.1.2

Divide $y$ by $1$ .

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

Step 3.9.3

Simplify the right side.

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Step 3.9.3.1

Move the negative in front of the fraction.

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

Step 4

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

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

Step 5

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

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Step 5.1

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

Step 5.2

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

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Step 5.2.1

Set up the composite result function.

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

Step 5.2.2

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

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

Step 5.2.3

Simplify the denominator.

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Step 5.2.3.1

Add $- 2$ and $2$ .

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

Step 5.2.3.2

Add $3^{x + 1}$ and $0$ .

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

Step 5.3

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

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Step 5.3.1

Set up the composite result function.

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

Step 5.3.2

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

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

Step 5.3.3

Simplify each term.

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Step 5.3.3.1

Write $1$ as a fraction with a common denominator.

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

Step 5.3.3.2

Combine the numerators over the common denominator.

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

Step 5.4

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

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