Precalculus Examples

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

Step 1

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

$y = 3^{x - 1}$

Step 2

Interchange the variables.

$x = 3^{y - 1}$

Step 3

Solve for $y$ .

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Step 3.1

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

$3^{y - 1} = x$

Step 3.2

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

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

Step 3.3

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

$(y - 1) \ln (3) = \ln (x)$

Step 3.4

Simplify the left side.

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Step 3.4.1

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

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Step 3.4.1.1

Apply the distributive property.

$y \ln (3) - 1 \ln (3) = \ln (x)$

Step 3.4.1.2

Rewrite $- 1 \ln (3)$ as $- \ln (3)$ .

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

Step 3.5

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

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

Step 3.6

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

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Step 3.6.1

Add $\ln (3)$ to both sides of the equation.

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

Step 3.6.2

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

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

Step 3.7

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

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Step 3.7.1

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

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

Step 3.7.2

Simplify the left side.

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Step 3.7.2.1

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

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Step 3.7.2.1.1

Cancel the common factor.

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

Step 3.7.2.1.2

Divide $y$ by $1$ .

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

Step 3.7.3

Simplify the right side.

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Step 3.7.3.1

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

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Step 3.7.3.1.1

Cancel the common factor.

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

Step 3.7.3.1.2

Rewrite the expression.

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

Step 4

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

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

Step 5

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

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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 $f^{- 1} \cdot 3^{x - 1}$ by substituting in the value of $f$ into $f^{- 1}$ .

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

Step 5.2.3

Simplify each term.

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Step 5.2.3.1

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

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

Step 5.2.3.2

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

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Step 5.2.3.2.1

Cancel the common factor.

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

Step 5.2.3.2.2

Divide $x - 1$ by $1$ .

$f^{- 1} \cdot 3^{x - 1} = 1 + x - 1$

Step 5.2.4

Combine the opposite terms in $1 + x - 1$ .

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Step 5.2.4.1

Subtract $1$ from $1$ .

$f^{- 1} \cdot 3^{x - 1} = x + 0$

Step 5.2.4.2

Add $x$ and $0$ .

$f^{- 1} \cdot 3^{x - 1} = x$

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

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

Step 5.3.3

Combine the opposite terms in $(1 + \frac{\ln (x)}{\ln (3)}) - 1$ .

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Step 5.3.3.1

Subtract $1$ from $1$ .

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

Step 5.3.3.2

Add $0$ and $\frac{\ln (x)}{\ln (3)}$ .

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

Step 5.3.4

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

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

Step 5.3.5

Exponentiation and log are inverse functions.

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

Step 5.4

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

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