Precalculus Examples

$y = 10^{x}$

Step 1

Interchange the variables.

$x = 10^{y}$

Step 2

Solve for $y$ .

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Step 2.1

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

$10^{y} = x$

Step 2.2

Take the base $10$ logarithm of both sides of the equation to remove the variable from the exponent.

$\log (10^{y}) = \log (x)$

Step 2.3

Expand the left side.

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Step 2.3.1

Expand $\log (10^{y})$ by moving $y$ outside the logarithm.

$y \log (10) = \log (x)$

Step 2.3.2

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

$y \cdot 1 = \log (x)$

Step 2.3.3

Multiply $y$ by $1$ .

$y = \log (x)$

Step 3

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

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

Step 4

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

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Step 4.1

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

Step 4.2

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

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Step 4.2.1

Set up the composite result function.

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

Step 4.2.2

Evaluate $f^{- 1} \cdot 10^{x}$ by substituting in the value of $f$ into $f^{- 1}$ .

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

Step 4.2.3

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

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

Step 4.2.4

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

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

Step 4.2.5

Multiply $x$ by $1$ .

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

Step 4.3

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

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Step 4.3.1

Set up the composite result function.

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

Step 4.3.2

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

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

Step 4.3.3

Exponentiation and log are inverse functions.

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

Step 4.4

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

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