Precalculus Examples

$f (x) = e^{3 x - 5} - 2$

Step 1

Write $f (x) = e^{3 x - 5} - 2$ as an equation.

$y = e^{3 x - 5} - 2$

Step 2

Interchange the variables.

$x = e^{3 y - 5} - 2$

Step 3

Solve for $y$ .

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Step 3.1

Rewrite the equation as $e^{3 y - 5} - 2 = x$ .

$e^{3 y - 5} - 2 = x$

Step 3.2

Add $2$ to both sides of the equation.

$e^{3 y - 5} = x + 2$

Step 3.3

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

$\ln (e^{3 y - 5}) = \ln (x + 2)$

Step 3.4

Expand the left side.

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Step 3.4.1

Expand $\ln (e^{3 y - 5})$ by moving $3 y - 5$ outside the logarithm.

$(3 y - 5) \ln (e) = \ln (x + 2)$

Step 3.4.2

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

$(3 y - 5) \cdot 1 = \ln (x + 2)$

Step 3.4.3

Multiply $3 y - 5$ by $1$ .

$3 y - 5 = \ln (x + 2)$

Step 3.5

Add $5$ to both sides of the equation.

$3 y = \ln (x + 2) + 5$

Step 3.6

Divide each term in $3 y = \ln (x + 2) + 5$ by $3$ and simplify.

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Step 3.6.1

Divide each term in $3 y = \ln (x + 2) + 5$ by $3$ .

$\frac{3 y}{3} = \frac{\ln (x + 2)}{3} + \frac{5}{3}$

Step 3.6.2

Simplify the left side.

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Step 3.6.2.1

Cancel the common factor of $3$ .

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Step 3.6.2.1.1

Cancel the common factor.

$\frac{3 y}{3} = \frac{\ln (x + 2)}{3} + \frac{5}{3}$

Step 3.6.2.1.2

Divide $y$ by $1$ .

$y = \frac{\ln (x + 2)}{3} + \frac{5}{3}$

Step 4

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

$f^{- 1} (x) = \frac{\ln (x + 2)}{3} + \frac{5}{3}$

Step 5

Verify if $f^{- 1} (x) = \frac{\ln (x + 2)}{3} + \frac{5}{3}$ is the inverse of $f (x) = e^{3 x - 5} - 2$ .

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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 $f^{- 1} (e^{3 x - 5} - 2)$ by substituting in the value of $f$ into $f^{- 1}$ .

$f^{- 1} (e^{3 x - 5} - 2) = \frac{\ln ((e^{3 x - 5} - 2) + 2)}{3} + \frac{5}{3}$

Step 5.2.3

Simplify terms.

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Step 5.2.3.1

Combine the opposite terms in $\frac{\ln ((e^{3 x - 5} - 2) + 2)}{3} + \frac{5}{3}$ .

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Step 5.2.3.1.1

Add $- 2$ and $2$ .

$f^{- 1} (e^{3 x - 5} - 2) = \frac{\ln (e^{3 x - 5} + 0)}{3} + \frac{5}{3}$

Step 5.2.3.1.2

Add $e^{3 x - 5}$ and $0$ .

$f^{- 1} (e^{3 x - 5} - 2) = \frac{\ln (e^{3 x - 5})}{3} + \frac{5}{3}$

Step 5.2.3.2

Combine the numerators over the common denominator.

$f^{- 1} (e^{3 x - 5} - 2) = \frac{\ln (e^{3 x - 5}) + 5}{3}$

Step 5.2.4

Simplify each term.

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Step 5.2.4.1

Use logarithm rules to move $3 x - 5$ out of the exponent.

$f^{- 1} (e^{3 x - 5} - 2) = \frac{(3 x - 5) \ln (e) + 5}{3}$

Step 5.2.4.2

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

$f^{- 1} (e^{3 x - 5} - 2) = \frac{(3 x - 5) \cdot 1 + 5}{3}$

Step 5.2.4.3

Multiply $3 x - 5$ by $1$ .

$f^{- 1} (e^{3 x - 5} - 2) = \frac{3 x - 5 + 5}{3}$

Step 5.2.5

Simplify terms.

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Step 5.2.5.1

Combine the opposite terms in $3 x - 5 + 5$ .

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Step 5.2.5.1.1

Add $- 5$ and $5$ .

$f^{- 1} (e^{3 x - 5} - 2) = \frac{3 x + 0}{3}$

Step 5.2.5.1.2

Add $3 x$ and $0$ .

$f^{- 1} (e^{3 x - 5} - 2) = \frac{3 x}{3}$

Step 5.2.5.2

Cancel the common factor of $3$ .

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Step 5.2.5.2.1

Cancel the common factor.

$f^{- 1} (e^{3 x - 5} - 2) = \frac{3 x}{3}$

Step 5.2.5.2.2

Divide $x$ by $1$ .

$f^{- 1} (e^{3 x - 5} - 2) = x$

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 $f (\frac{\ln (x + 2)}{3} + \frac{5}{3})$ by substituting in the value of $f^{- 1}$ into $f$ .

$f (\frac{\ln (x + 2)}{3} + \frac{5}{3}) = e^{3 (\frac{\ln (x + 2)}{3} + \frac{5}{3}) - 5} - 2$

Step 5.3.3

Simplify each term.

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Step 5.3.3.1

Simplify each term.

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Step 5.3.3.1.1

Simplify each term.

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Step 5.3.3.1.1.1

Rewrite $\frac{\ln (x + 2)}{3}$ as $\frac{1}{3} \ln (x + 2)$ .

$f (\frac{\ln (x + 2)}{3} + \frac{5}{3}) = e^{3 (\frac{1}{3} \cdot \ln (x + 2) + \frac{5}{3}) - 5} - 2$

Step 5.3.3.1.1.2

Simplify $\frac{1}{3} \ln (x + 2)$ by moving $\frac{1}{3}$ inside the logarithm.

$f (\frac{\ln (x + 2)}{3} + \frac{5}{3}) = e^{3 (\ln ({(x + 2)}^{\frac{1}{3}}) + \frac{5}{3}) - 5} - 2$

Step 5.3.3.1.2

To write $\ln ({(x + 2)}^{\frac{1}{3}})$ as a fraction with a common denominator, multiply by $\frac{3}{3}$ .

$f (\frac{\ln (x + 2)}{3} + \frac{5}{3}) = e^{3 (\ln ({(x + 2)}^{\frac{1}{3}}) \cdot \frac{3}{3} + \frac{5}{3}) - 5} - 2$

Step 5.3.3.1.3

Combine $\ln ({(x + 2)}^{\frac{1}{3}})$ and $\frac{3}{3}$ .

$f (\frac{\ln (x + 2)}{3} + \frac{5}{3}) = e^{3 (\frac{\ln ({(x + 2)}^{\frac{1}{3}}) \cdot 3}{3} + \frac{5}{3}) - 5} - 2$

Step 5.3.3.1.4

Combine the numerators over the common denominator.

$f (\frac{\ln (x + 2)}{3} + \frac{5}{3}) = e^{3 (\frac{\ln ({(x + 2)}^{\frac{1}{3}}) \cdot 3 + 5}{3}) - 5} - 2$

Step 5.3.3.1.5

Cancel the common factor of $3$ .

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Step 5.3.3.1.5.1

Simplify the numerator.

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Step 5.3.3.1.5.1.1

Multiply $\ln ({(x + 2)}^{\frac{1}{3}}) \cdot 3$ .

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Step 5.3.3.1.5.1.1.1

Reorder $\ln ({(x + 2)}^{\frac{1}{3}})$ and $3$ .

$f (\frac{\ln (x + 2)}{3} + \frac{5}{3}) = e^{3 (\frac{3 \cdot \ln ({(x + 2)}^{\frac{1}{3}}) + 5}{3}) - 5} - 2$

Step 5.3.3.1.5.1.1.2

Simplify $3 \cdot \ln ({(x + 2)}^{\frac{1}{3}})$ by moving $3$ inside the logarithm.

$f (\frac{\ln (x + 2)}{3} + \frac{5}{3}) = e^{3 (\frac{\ln ({({(x + 2)}^{\frac{1}{3}})}^{3}) + 5}{3}) - 5} - 2$

Step 5.3.3.1.5.1.2

Multiply the exponents in ${({(x + 2)}^{\frac{1}{3}})}^{3}$ .

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Step 5.3.3.1.5.1.2.1

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$f (\frac{\ln (x + 2)}{3} + \frac{5}{3}) = e^{3 (\frac{\ln ({(x + 2)}^{\frac{1}{3} \cdot 3}) + 5}{3}) - 5} - 2$

Step 5.3.3.1.5.1.2.2

Cancel the common factor of $3$ .

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Step 5.3.3.1.5.1.2.2.1

Cancel the common factor.

$f (\frac{\ln (x + 2)}{3} + \frac{5}{3}) = e^{3 (\frac{\ln ({(x + 2)}^{\frac{1}{3} \cdot 3}) + 5}{3}) - 5} - 2$

Step 5.3.3.1.5.1.2.2.2

Rewrite the expression.

$f (\frac{\ln (x + 2)}{3} + \frac{5}{3}) = e^{3 (\frac{\ln ({(x + 2)}^{1}) + 5}{3}) - 5} - 2$

Step 5.3.3.1.5.1.3

Simplify.

$f (\frac{\ln (x + 2)}{3} + \frac{5}{3}) = e^{3 (\frac{\ln (x + 2) + 5}{3}) - 5} - 2$

Step 5.3.3.1.5.2

Cancel the common factor.

$f (\frac{\ln (x + 2)}{3} + \frac{5}{3}) = e^{3 (\frac{\ln (x + 2) + 5}{3}) - 5} - 2$

Step 5.3.3.1.5.3

Rewrite the expression.

$f (\frac{\ln (x + 2)}{3} + \frac{5}{3}) = e^{\ln (x + 2) + 5 - 5} - 2$

Step 5.3.3.2

Combine the opposite terms in $\ln (x + 2) + 5 - 5$ .

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Step 5.3.3.2.1

Subtract $5$ from $5$ .

$f (\frac{\ln (x + 2)}{3} + \frac{5}{3}) = e^{\ln (x + 2) + 0} - 2$

Step 5.3.3.2.2

Add $\ln (x + 2)$ and $0$ .

$f (\frac{\ln (x + 2)}{3} + \frac{5}{3}) = e^{\ln (x + 2)} - 2$

Step 5.3.3.3

Exponentiation and log are inverse functions.

$f (\frac{\ln (x + 2)}{3} + \frac{5}{3}) = x + 2 - 2$

Step 5.3.4

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

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Step 5.3.4.1

Subtract $2$ from $2$ .

$f (\frac{\ln (x + 2)}{3} + \frac{5}{3}) = x + 0$

Step 5.3.4.2

Add $x$ and $0$ .

$f (\frac{\ln (x + 2)}{3} + \frac{5}{3}) = x$

Step 5.4

Since $f^{- 1} (f (x)) = x$ and $f (f^{- 1} (x)) = x$ , then $f^{- 1} (x) = \frac{\ln (x + 2)}{3} + \frac{5}{3}$ is the inverse of $f (x) = e^{3 x - 5} - 2$ .

$f^{- 1} (x) = \frac{\ln (x + 2)}{3} + \frac{5}{3}$