Algebra Examples

$y = \frac{4^{x}}{2}$

Step 1

Interchange the variables.

$x = \frac{4^{y}}{2}$

Step 2

Solve for $y$ .

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Step 2.1

Rewrite the equation as $\frac{4^{y}}{2} = x$ .

$\frac{4^{y}}{2} = x$

Step 2.2

Multiply both sides of the equation by $2$ .

$2 \frac{4^{y}}{2} = 2 x$

Step 2.3

Simplify the left side.

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Step 2.3.1

Cancel the common factor of $2$ .

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Step 2.3.1.1

Cancel the common factor.

$2 \frac{4^{y}}{2} = 2 x$

Step 2.3.1.2

Rewrite the expression.

$4^{y} = 2 x$

Step 2.4

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

$\ln (4^{y}) = \ln (2 x)$

Step 2.5

Expand $\ln (4^{y})$ by moving $y$ outside the logarithm.

$y \ln (4) = \ln (2 x)$

Step 2.6

Divide each term in $y \ln (4) = \ln (2 x)$ by $\ln (4)$ and simplify.

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Step 2.6.1

Divide each term in $y \ln (4) = \ln (2 x)$ by $\ln (4)$ .

$\frac{y \ln (4)}{\ln (4)} = \frac{\ln (2 x)}{\ln (4)}$

Step 2.6.2

Simplify the left side.

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Step 2.6.2.1

Cancel the common factor of $\ln (4)$ .

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Step 2.6.2.1.1

Cancel the common factor.

$\frac{y \ln (4)}{\ln (4)} = \frac{\ln (2 x)}{\ln (4)}$

Step 2.6.2.1.2

Divide $y$ by $1$ .

$y = \frac{\ln (2 x)}{\ln (4)}$

Step 3

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

$f^{- 1} (x) = \frac{\ln (2 x)}{\ln (4)}$

Step 4

Verify if $f^{- 1} (x) = \frac{\ln (2 x)}{\ln (4)}$ is the inverse of $f (x) = \frac{4^{x}}{2}$ .

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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 $f^{- 1} (\frac{4^{x}}{2})$ by substituting in the value of $f$ into $f^{- 1}$ .

$f^{- 1} (\frac{4^{x}}{2}) = \frac{\ln (2 (\frac{4^{x}}{2}))}{\ln (4)}$

Step 4.2.3

Cancel the common factor of $2$ .

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Step 4.2.3.1

Cancel the common factor.

$f^{- 1} (\frac{4^{x}}{2}) = \frac{\ln (2 (\frac{4^{x}}{2}))}{\ln (4)}$

Step 4.2.3.2

Rewrite the expression.

$f^{- 1} (\frac{4^{x}}{2}) = \frac{\ln (4^{x})}{\ln (4)}$

Step 4.2.4

Expand $\ln (4^{x})$ by moving $x$ outside the logarithm.

$f^{- 1} (\frac{4^{x}}{2}) = \frac{x \ln (4)}{\ln (4)}$

Step 4.2.5

Cancel the common factor of $\ln (4)$ .

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Step 4.2.5.1

Cancel the common factor.

$f^{- 1} (\frac{4^{x}}{2}) = \frac{x \ln (4)}{\ln (4)}$

Step 4.2.5.2

Divide $x$ by $1$ .

$f^{- 1} (\frac{4^{x}}{2}) = x$

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 $f (\frac{\ln (2 x)}{\ln (4)})$ by substituting in the value of $f^{- 1}$ into $f$ .

$f (\frac{\ln (2 x)}{\ln (4)}) = \frac{4^{\frac{\ln (2 x)}{\ln (4)}}}{2}$

Step 4.3.3

Simplify the numerator.

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Step 4.3.3.1

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

$f (\frac{\ln (2 x)}{\ln (4)}) = \frac{4^{\log_{4} (2 x)}}{2}$

Step 4.3.3.2

Exponentiation and log are inverse functions.

$f (\frac{\ln (2 x)}{\ln (4)}) = \frac{2 x}{2}$

Step 4.3.4

Cancel the common factor of $2$ .

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Step 4.3.4.1

Cancel the common factor.

$f (\frac{\ln (2 x)}{\ln (4)}) = \frac{2 x}{2}$

Step 4.3.4.2

Divide $x$ by $1$ .

$f (\frac{\ln (2 x)}{\ln (4)}) = x$

Step 4.4

Since $f^{- 1} (f (x)) = x$ and $f (f^{- 1} (x)) = x$ , then $f^{- 1} (x) = \frac{\ln (2 x)}{\ln (4)}$ is the inverse of $f (x) = \frac{4^{x}}{2}$ .

$f^{- 1} (x) = \frac{\ln (2 x)}{\ln (4)}$