Precalculus Examples

$h (x) = \log_{7} (x)$

Step 1

Write $h (x) = \log_{7} (x)$ as an equation.

$y = \log_{7} (x)$

Step 2

Interchange the variables.

$x = \log_{7} (y)$

Step 3

Solve for $y$ .

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Step 3.1

Rewrite the equation as $\log_{7} (y) = x$ .

$\log_{7} (y) = x$

Step 3.2

Rewrite $\log_{7} (y) = x$ 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 $b^{y} = x$ .

$7^{x} = y$

Step 3.3

Rewrite the equation as $y = 7^{x}$ .

$y = 7^{x}$

Step 4

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

$h^{- 1} (x) = 7^{x}$

Step 5

Verify if $h^{- 1} (x) = 7^{x}$ is the inverse of $h (x) = \log_{7} (x)$ .

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Step 5.1

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

Step 5.2

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

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Step 5.2.1

Set up the composite result function.

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

Step 5.2.2

Evaluate $h^{- 1} (\log_{7} (x))$ by substituting in the value of $h$ into $h^{- 1}$ .

$h^{- 1} (\log_{7} (x)) = 7^{\log_{7} (x)}$

Step 5.2.3

Exponentiation and log are inverse functions.

$h^{- 1} (\log_{7} (x)) = x$

Step 5.3

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

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Step 5.3.1

Set up the composite result function.

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

Step 5.3.2

Evaluate $h (7^{x})$ by substituting in the value of $h^{- 1}$ into $h$ .

$h (7^{x}) = \log_{7} (7^{x})$

Step 5.3.3

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

$h (7^{x}) = x \log_{7} (7)$

Step 5.3.4

Logarithm base $7$ of $7$ is $1$ .

$h (7^{x}) = x \cdot 1$

Step 5.3.5

Multiply $x$ by $1$ .

$h (7^{x}) = x$

Step 5.4

Since $h^{- 1} (h (x)) = x$ and $h (h^{- 1} (x)) = x$ , then $h^{- 1} (x) = 7^{x}$ is the inverse of $h (x) = \log_{7} (x)$ .

$h^{- 1} (x) = 7^{x}$