Precalculus Examples

$\ln (x - 2)$

Step 1

Interchange the variables.

$x = \ln (y - 2)$

Step 2

Solve for $y$ .

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

$\ln (y - 2) = x$

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

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

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

$e^{x} = y - 2$

Solve for $y$ .

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

$y - 2 = e^{x}$

Add $2$ to both sides of the equation.

$y = e^{x} + 2$

Step 3

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

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

Step 4

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

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

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

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Set up the composite result function.

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

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

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

Exponentiation and log are inverse functions.

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

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

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Add $- 2$ and $2$ .

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

Add $x$ and $0$ .

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

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

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Set up the composite result function.

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

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

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

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

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Subtract $2$ from $2$ .

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

Add $e^{x}$ and $0$ .

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

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

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

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

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

Multiply $x$ by $1$ .

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

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

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