Precalculus Examples

$y = 5^{x}$

Step 1

Interchange the variables.

$x = 5^{y}$

Step 2

Solve for $y$ .

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Step 2.1

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

$5^{y} = x$

Step 2.2

Take the natural logarithm of both sides of the equation to remove the variable from the exponent.

$\ln (5^{y}) = \ln (x)$

Step 2.3

Expand $\ln (5^{y})$ by moving $y$ outside the logarithm.

$y \ln (5) = \ln (x)$

Step 2.4

Divide each term in $y \ln (5) = \ln (x)$ by $\ln (5)$ and simplify.

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Step 2.4.1

Divide each term in $y \ln (5) = \ln (x)$ by $\ln (5)$ .

$\frac{y \ln (5)}{\ln (5)} = \frac{\ln (x)}{\ln (5)}$

Step 2.4.2

Simplify the left side.

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Step 2.4.2.1

Cancel the common factor of $\ln (5)$ .

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Step 2.4.2.1.1

Cancel the common factor.

$\frac{y \ln (5)}{\ln (5)} = \frac{\ln (x)}{\ln (5)}$

Step 2.4.2.1.2

Divide $y$ by $1$ .

$y = \frac{\ln (x)}{\ln (5)}$

Step 3

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

$f^{- 1} (x) = \frac{\ln (x)}{\ln (5)}$

Step 4

Verify if $f^{- 1} (x) = \frac{\ln (x)}{\ln (5)}$ is the inverse of $f (x) = 5^{x}$ .

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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 $f^{- 1} \cdot 5^{x}$ by substituting in the value of $f$ into $f^{- 1}$ .

$f^{- 1} \cdot 5^{x} = \frac{\ln (5^{x})}{\ln (5)}$

Step 4.2.3

Expand $\ln (5^{x})$ by moving $x$ outside the logarithm.

$f^{- 1} \cdot 5^{x} = \frac{x \ln (5)}{\ln (5)}$

Step 4.2.4

Cancel the common factor of $\ln (5)$ .

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Step 4.2.4.1

Cancel the common factor.

$f^{- 1} \cdot 5^{x} = \frac{x \ln (5)}{\ln (5)}$

Step 4.2.4.2

Divide $x$ by $1$ .

$f^{- 1} \cdot 5^{x} = x$

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 $f (\frac{\ln (x)}{\ln (5)})$ by substituting in the value of $f^{- 1}$ into $f$ .

$f (\frac{\ln (x)}{\ln (5)}) = 5^{\frac{\ln (x)}{\ln (5)}}$

Step 4.3.3

Use the change of base rule $\frac{\log_{b} (x)}{\log_{b} (a)} = \log_{a} (x)$ .

$f (\frac{\ln (x)}{\ln (5)}) = 5^{\log_{5} (x)}$

Step 4.3.4

Exponentiation and log are inverse functions.

$f (\frac{\ln (x)}{\ln (5)}) = x$

Step 4.4

Since $f^{- 1} (f (x)) = x$ and $f (f^{- 1} (x)) = x$ , then $f^{- 1} (x) = \frac{\ln (x)}{\ln (5)}$ is the inverse of $f (x) = 5^{x}$ .

$f^{- 1} (x) = \frac{\ln (x)}{\ln (5)}$