Precalculus Examples

$y = 8^{2 x}$

Step 1

Interchange the variables.

$x = 8^{2 y}$

Step 2

Solve for $y$ .

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Step 2.1

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

$8^{2 y} = x$

Step 2.2

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

$\ln (8^{2 y}) = \ln (x)$

Step 2.3

Expand $\ln (8^{2 y})$ by moving $2 y$ outside the logarithm.

$2 y \ln (8) = \ln (x)$

Step 2.4

Divide each term in $2 y \ln (8) = \ln (x)$ by $2 \ln (8)$ and simplify.

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Step 2.4.1

Divide each term in $2 y \ln (8) = \ln (x)$ by $2 \ln (8)$ .

$\frac{2 y \ln (8)}{2 \ln (8)} = \frac{\ln (x)}{2 \ln (8)}$

Step 2.4.2

Simplify the left side.

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Step 2.4.2.1

Cancel the common factor of $2$ .

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Step 2.4.2.1.1

Cancel the common factor.

$\frac{2 y \ln (8)}{2 \ln (8)} = \frac{\ln (x)}{2 \ln (8)}$

Step 2.4.2.1.2

Rewrite the expression.

$\frac{y \ln (8)}{\ln (8)} = \frac{\ln (x)}{2 \ln (8)}$

Step 2.4.2.2

Cancel the common factor of $\ln (8)$ .

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Step 2.4.2.2.1

Cancel the common factor.

$\frac{y \ln (8)}{\ln (8)} = \frac{\ln (x)}{2 \ln (8)}$

Step 2.4.2.2.2

Divide $y$ by $1$ .

$y = \frac{\ln (x)}{2 \ln (8)}$

Step 3

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

$f^{- 1} (x) = \frac{\ln (x)}{2 \ln (8)}$

Step 4

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

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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 $f^{- 1} \cdot 8^{2 x}$ by substituting in the value of $f$ into $f^{- 1}$ .

$f^{- 1} \cdot 8^{2 x} = \frac{\ln (8^{2 x})}{2 \ln (8)}$

Step 4.2.3

Expand $\ln (8^{2 x})$ by moving $2 x$ outside the logarithm.

$f^{- 1} \cdot 8^{2 x} = \frac{2 x \ln (8)}{2 \ln (8)}$

Step 4.2.4

Cancel the common factor of $2$ .

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Step 4.2.4.1

Cancel the common factor.

$f^{- 1} \cdot 8^{2 x} = \frac{2 x \ln (8)}{2 \ln (8)}$

Step 4.2.4.2

Rewrite the expression.

$f^{- 1} \cdot 8^{2 x} = \frac{x \ln (8)}{\ln (8)}$

Step 4.2.5

Cancel the common factor of $\ln (8)$ .

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Step 4.2.5.1

Cancel the common factor.

$f^{- 1} \cdot 8^{2 x} = \frac{x \ln (8)}{\ln (8)}$

Step 4.2.5.2

Divide $x$ by $1$ .

$f^{- 1} \cdot 8^{2 x} = x$

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

$f (\frac{\ln (x)}{2 \ln (8)}) = 8^{2 (\frac{\ln (x)}{2 \ln (8)})}$

Step 4.3.3

Simplify $2 \ln (8)$ by moving $2$ inside the logarithm.

$f (\frac{\ln (x)}{2 \ln (8)}) = 8^{2 (\frac{\ln (x)}{\ln (8^{2})})}$

Step 4.3.4

Raise $8$ to the power of $2$ .

$f (\frac{\ln (x)}{2 \ln (8)}) = 8^{2 (\frac{\ln (x)}{\ln (64)})}$

Step 4.3.5

Multiply $2 \frac{\ln (x)}{\ln (64)}$ .

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Step 4.3.5.1

Combine $2$ and $\frac{\ln (x)}{\ln (64)}$ .

$f (\frac{\ln (x)}{2 \ln (8)}) = 8^{\frac{2 \ln (x)}{\ln (64)}}$

Step 4.3.5.2

Simplify $2 \ln (x)$ by moving $2$ inside the logarithm.

$f (\frac{\ln (x)}{2 \ln (8)}) = 8^{\frac{\ln (x^{2})}{\ln (64)}}$

Step 4.4

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

$f^{- 1} (x) = \frac{\ln (x)}{2 \ln (8)}$