Finite Math Examples

$f (x) = \log_{2} (x - 8) + 5$

Step 1

Write $f (x) = \log_{2} (x - 8) + 5$ as an equation.

$y = \log_{2} (x - 8) + 5$

Step 2

Interchange the variables.

$x = \log_{2} (y - 8) + 5$

Step 3

Solve for $y$ .

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Step 3.1

Rewrite the equation as $\log_{2} (y - 8) + 5 = x$ .

$\log_{2} (y - 8) + 5 = x$

Step 3.2

Subtract $5$ from both sides of the equation.

$\log_{2} (y - 8) = x - 5$

Step 3.3

Rewrite $\log_{2} (y - 8) = x - 5$ 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 $b^{y} = x$ .

$2^{x - 5} = y - 8$

Step 3.4

Solve for $y$ .

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Step 3.4.1

Rewrite the equation as $y - 8 = 2^{x - 5}$ .

$y - 8 = 2^{x - 5}$

Step 3.4.2

Add $8$ to both sides of the equation.

$y = 2^{x - 5} + 8$

Step 4

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

$f^{- 1} (x) = 2^{x - 5} + 8$

Step 5

Verify if $f^{- 1} (x) = 2^{x - 5} + 8$ is the inverse of $f (x) = \log_{2} (x - 8) + 5$ .

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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 $f^{- 1} (\log_{2} (x - 8) + 5)$ by substituting in the value of $f$ into $f^{- 1}$ .

$f^{- 1} (\log_{2} (x - 8) + 5) = 2^{(\log_{2} (x - 8) + 5) - 5} + 8$

Step 5.2.3

Combine the opposite terms in $2^{(\log_{2} (x - 8) + 5) - 5} + 8$ .

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Step 5.2.3.1

Subtract $5$ from $5$ .

$f^{- 1} (\log_{2} (x - 8) + 5) = 2^{\log_{2} (x - 8) + 0} + 8$

Step 5.2.3.2

Add $\log_{2} (x - 8)$ and $0$ .

$f^{- 1} (\log_{2} (x - 8) + 5) = 2^{\log_{2} (x - 8)} + 8$

Step 5.2.4

Exponentiation and log are inverse functions.

$f^{- 1} (\log_{2} (x - 8) + 5) = x - 8 + 8$

Step 5.2.5

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

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Step 5.2.5.1

Add $- 8$ and $8$ .

$f^{- 1} (\log_{2} (x - 8) + 5) = x + 0$

Step 5.2.5.2

Add $x$ and $0$ .

$f^{- 1} (\log_{2} (x - 8) + 5) = x$

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 $f (2^{x - 5} + 8)$ by substituting in the value of $f^{- 1}$ into $f$ .

$f (2^{x - 5} + 8) = \log_{2} ((2^{x - 5} + 8) - 8) + 5$

Step 5.3.3

Combine the opposite terms in $\log_{2} ((2^{x - 5} + 8) - 8) + 5$ .

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Step 5.3.3.1

Subtract $8$ from $8$ .

$f (2^{x - 5} + 8) = \log_{2} (2^{x - 5} + 0) + 5$

Step 5.3.3.2

Add $2^{x - 5}$ and $0$ .

$f (2^{x - 5} + 8) = \log_{2} (2^{x - 5}) + 5$

Step 5.3.4

Simplify each term.

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Step 5.3.4.1

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

$f (2^{x - 5} + 8) = (x - 5) \log_{2} (2) + 5$

Step 5.3.4.2

Logarithm base $2$ of $2$ is $1$ .

$f (2^{x - 5} + 8) = (x - 5) \cdot 1 + 5$

Step 5.3.4.3

Multiply $x - 5$ by $1$ .

$f (2^{x - 5} + 8) = x - 5 + 5$

Step 5.3.5

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

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Step 5.3.5.1

Add $- 5$ and $5$ .

$f (2^{x - 5} + 8) = x + 0$

Step 5.3.5.2

Add $x$ and $0$ .

$f (2^{x - 5} + 8) = x$

Step 5.4

Since $f^{- 1} (f (x)) = x$ and $f (f^{- 1} (x)) = x$ , then $f^{- 1} (x) = 2^{x - 5} + 8$ is the inverse of $f (x) = \log_{2} (x - 8) + 5$ .

$f^{- 1} (x) = 2^{x - 5} + 8$