Precalculus Examples

$f (x) = \log (x + 8)$

Step 1

Write $f (x) = \log (x + 8)$ as an equation.

$y = \log (x + 8)$

Step 2

Interchange the variables.

$x = \log (y + 8)$

Step 3

Solve for $y$ .

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Step 3.1

Rewrite the equation as $\log (y + 8) = x$ .

$\log (y + 8) = x$

Step 3.2

Rewrite $\log (y + 8) = x$ 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 $b^{y} = x$ .

$10^{x} = y + 8$

Step 3.3

Solve for $y$ .

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Step 3.3.1

Rewrite the equation as $y + 8 = 10^{x}$ .

$y + 8 = 10^{x}$

Step 3.3.2

Subtract $8$ from both sides of the equation.

$y = 10^{x} - 8$

Step 4

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

$f^{- 1} (x) = 10^{x} - 8$

Step 5

Verify if $f^{- 1} (x) = 10^{x} - 8$ is the inverse of $f (x) = \log (x + 8)$ .

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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 $f^{- 1} (\log (x + 8))$ by substituting in the value of $f$ into $f^{- 1}$ .

$f^{- 1} (\log (x + 8)) = 10^{\log (x + 8)} - 8$

Step 5.2.3

Exponentiation and log are inverse functions.

$f^{- 1} (\log (x + 8)) = x + 8 - 8$

Step 5.2.4

Combine the opposite terms in $x + 8 - 8$ .

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Step 5.2.4.1

Subtract $8$ from $8$ .

$f^{- 1} (\log (x + 8)) = x + 0$

Step 5.2.4.2

Add $x$ and $0$ .

$f^{- 1} (\log (x + 8)) = x$

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 $f (10^{x} - 8)$ by substituting in the value of $f^{- 1}$ into $f$ .

$f (10^{x} - 8) = \log ((10^{x} - 8) + 8)$

Step 5.3.3

Combine the opposite terms in $(10^{x} - 8) + 8$ .

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Step 5.3.3.1

Add $- 8$ and $8$ .

$f (10^{x} - 8) = \log (10^{x} + 0)$

Step 5.3.3.2

Add $10^{x}$ and $0$ .

$f (10^{x} - 8) = \log (10^{x})$

Step 5.3.4

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

$f (10^{x} - 8) = x \log (10)$

Step 5.3.5

Logarithm base $10$ of $10$ is $1$ .

$f (10^{x} - 8) = x \cdot 1$

Step 5.3.6

Multiply $x$ by $1$ .

$f (10^{x} - 8) = x$

Step 5.4

Since $f^{- 1} (f (x)) = x$ and $f (f^{- 1} (x)) = x$ , then $f^{- 1} (x) = 10^{x} - 8$ is the inverse of $f (x) = \log (x + 8)$ .

$f^{- 1} (x) = 10^{x} - 8$