Precalculus Examples

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

Step 1

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

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

Step 2

Interchange the variables.

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

Step 3

Solve for $y$ .

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Step 3.1

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

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

Step 3.2

Add $2$ to both sides of the equation.

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

Step 3.3

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

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

Step 3.4

Rewrite $\ln (3 y) = x + 2$ 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 $b^{y} = x$ .

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

Step 3.5

Solve for $y$ .

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Step 3.5.1

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

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

Step 3.5.2

Divide each term in $3 y = e^{x + 2}$ by $3$ and simplify.

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Step 3.5.2.1

Divide each term in $3 y = e^{x + 2}$ by $3$ .

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

Step 3.5.2.2

Simplify the left side.

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Step 3.5.2.2.1

Cancel the common factor of $3$ .

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Step 3.5.2.2.1.1

Cancel the common factor.

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

Step 3.5.2.2.1.2

Divide $y$ by $1$ .

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

Step 4

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

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

Step 5

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

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

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

Step 5.2.3

Simplify the numerator.

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Step 5.2.3.1

Combine the opposite terms in $\ln (3 x) - 2 + 2$ .

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Step 5.2.3.1.1

Add $- 2$ and $2$ .

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

Step 5.2.3.1.2

Add $\ln (3 x)$ and $0$ .

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

Step 5.2.3.2

Exponentiation and log are inverse functions.

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

Step 5.2.4

Cancel the common factor of $3$ .

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Step 5.2.4.1

Cancel the common factor.

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

Step 5.2.4.2

Divide $x$ by $1$ .

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

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

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

Step 5.3.3

Simplify each term.

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Step 5.3.3.1

Cancel the common factor of $3$ .

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Step 5.3.3.1.1

Cancel the common factor.

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

Step 5.3.3.1.2

Rewrite the expression.

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

Step 5.3.3.2

Use logarithm rules to move $x + 2$ out of the exponent.

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

Step 5.3.3.3

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

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

Step 5.3.3.4

Multiply $x + 2$ by $1$ .

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

Step 5.3.4

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

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Step 5.3.4.1

Subtract $2$ from $2$ .

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

Step 5.3.4.2

Add $x$ and $0$ .

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

Step 5.4

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

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