Precalculus Examples

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

Step 1

Write $f (x) = \ln (x + 2) + \ln (3)$ as an equation.

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

Step 2

Interchange the variables.

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

Step 3

Solve for $y$ .

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Step 3.1

Rewrite the equation as $\ln (y + 2) + \ln (3) = x$ .

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

Step 3.2

Simplify the left side.

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Step 3.2.1

Use the product property of logarithms, $\log_{b} (x) + \log_{b} (y) = \log_{b} (x y)$ .

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

Step 3.2.2

Apply the distributive property.

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

Step 3.2.3

Simplify the expression.

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Step 3.2.3.1

Move $3$ to the left of $y$ .

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

Step 3.2.3.2

Multiply $2$ by $3$ .

$\ln (3 y + 6) = x$

Step 3.3

To solve for $y$ , rewrite the equation using properties of logarithms.

$e^{\ln (3 y + 6)} = e^{x}$

Step 3.4

Rewrite $\ln (3 y + 6) = x$ in exponential form using the definition of a logarithm. If $x$ and $b$ are positive real numbers and $b \neq 1$ , then $\log_{b} (x) = y$ is equivalent to $b^{y} = x$ .

$e^{x} = 3 y + 6$

Step 3.5

Solve for $y$ .

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Step 3.5.1

Rewrite the equation as $3 y + 6 = e^{x}$ .

$3 y + 6 = e^{x}$

Step 3.5.2

Subtract $6$ from both sides of the equation.

$3 y = e^{x} - 6$

Step 3.5.3

Divide each term in $3 y = e^{x} - 6$ by $3$ and simplify.

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Step 3.5.3.1

Divide each term in $3 y = e^{x} - 6$ by $3$ .

$\frac{3 y}{3} = \frac{e^{x}}{3} + \frac{- 6}{3}$

Step 3.5.3.2

Simplify the left side.

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Step 3.5.3.2.1

Cancel the common factor of $3$ .

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Step 3.5.3.2.1.1

Cancel the common factor.

$\frac{3 y}{3} = \frac{e^{x}}{3} + \frac{- 6}{3}$

Step 3.5.3.2.1.2

Divide $y$ by $1$ .

$y = \frac{e^{x}}{3} + \frac{- 6}{3}$

Step 3.5.3.3

Simplify the right side.

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Step 3.5.3.3.1

Divide $- 6$ by $3$ .

$y = \frac{e^{x}}{3} - 2$

Step 4

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

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

Step 5

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

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

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

Step 5.2.3

Use the product property of logarithms, $\log_{b} (x) + \log_{b} (y) = \log_{b} (x y)$ .

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

Step 5.2.4

Simplify each term.

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Step 5.2.4.1

Simplify the numerator.

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Step 5.2.4.1.1

Exponentiation and log are inverse functions.

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

Step 5.2.4.1.2

Apply the distributive property.

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

Step 5.2.4.1.3

Move $3$ to the left of $x$ .

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

Step 5.2.4.1.4

Multiply $2$ by $3$ .

$f^{- 1} (\ln (x + 2) + \ln (3)) = \frac{3 \cdot x + 6}{3} - 2$

Step 5.2.4.1.5

Factor $3$ out of $3 x + 6$ .

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Step 5.2.4.1.5.1

Factor $3$ out of $3 x$ .

$f^{- 1} (\ln (x + 2) + \ln (3)) = \frac{3 (x) + 6}{3} - 2$

Step 5.2.4.1.5.2

Factor $3$ out of $6$ .

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

Step 5.2.4.1.5.3

Factor $3$ out of $3 x + 3 \cdot 2$ .

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

Step 5.2.4.2

Cancel the common factor of $3$ .

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Step 5.2.4.2.1

Cancel the common factor.

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

Step 5.2.4.2.2

Divide $x + 2$ by $1$ .

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

Step 5.2.5

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

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Step 5.2.5.1

Subtract $2$ from $2$ .

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

Step 5.2.5.2

Add $x$ and $0$ .

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

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

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

Step 5.3.3

Combine the opposite terms in $\ln ((\frac{e^{x}}{3} - 2) + 2) + \ln (3)$ .

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Step 5.3.3.1

Add $- 2$ and $2$ .

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

Step 5.3.3.2

Add $\frac{e^{x}}{3}$ and $0$ .

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

Step 5.3.4

Use the product property of logarithms, $\log_{b} (x) + \log_{b} (y) = \log_{b} (x y)$ .

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

Step 5.3.5

Cancel the common factor of $3$ .

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Step 5.3.5.1

Cancel the common factor.

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

Step 5.3.5.2

Rewrite the expression.

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

Step 5.3.6

Use logarithm rules to move $x$ out of the exponent.

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

Step 5.3.7

The natural logarithm of $e$ is $1$ .

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

Step 5.3.8

Multiply $x$ by $1$ .

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

Step 5.4

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

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