Algebra Examples

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

Step 1

Interchange the variables.

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

Step 2

Solve for $y$ .

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Step 2.1

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

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

Step 2.2

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

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

Step 2.3

Expand the left side.

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Step 2.3.1

Expand $\ln (4^{\frac{y}{2}})$ by moving $\frac{y}{2}$ outside the logarithm.

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

Step 2.3.2

Combine $\frac{y}{2}$ and $\ln (4)$ .

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

Step 2.4

Multiply both sides of the equation by $\frac{2}{\ln (4)}$ .

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

Step 2.5

Simplify both sides of the equation.

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Step 2.5.1

Simplify the left side.

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Step 2.5.1.1

Simplify $\frac{2}{\ln (4)} \cdot \frac{y \ln (4)}{2}$ .

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Step 2.5.1.1.1

Cancel the common factor of $2$ .

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Step 2.5.1.1.1.1

Cancel the common factor.

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

Step 2.5.1.1.1.2

Rewrite the expression.

$\frac{1}{\ln (4)} (y \ln (4)) = \frac{2}{\ln (4)} \ln (x)$

Step 2.5.1.1.2

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

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Step 2.5.1.1.2.1

Factor $\ln (4)$ out of $y \ln (4)$ .

$\frac{1}{\ln (4)} (\ln (4) y) = \frac{2}{\ln (4)} \ln (x)$

Step 2.5.1.1.2.2

Cancel the common factor.

$\frac{1}{\ln (4)} (\ln (4) y) = \frac{2}{\ln (4)} \ln (x)$

Step 2.5.1.1.2.3

Rewrite the expression.

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

Step 2.5.2

Simplify the right side.

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Step 2.5.2.1

Simplify $\frac{2}{\ln (4)} \ln (x)$ .

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Step 2.5.2.1.1

Rewrite $\ln (4)$ as $\ln (2^{2})$ .

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

Step 2.5.2.1.2

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

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

Step 2.5.2.1.3

Cancel the common factor of $2$ .

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Step 2.5.2.1.3.1

Cancel the common factor.

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

Step 2.5.2.1.3.2

Rewrite the expression.

$y = \frac{1}{\ln (2)} \ln (x)$

Step 2.5.2.1.4

Combine $\frac{1}{\ln (2)}$ and $\ln (x)$ .

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

Step 3

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

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

Step 4

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

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

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

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

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

Step 4.3.3

Multiply the numerator by the reciprocal of the denominator.

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

Step 4.3.4

Multiply $\frac{\ln (x)}{\ln (2)} \cdot \frac{1}{2}$ .

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Step 4.3.4.1

Multiply $\frac{\ln (x)}{\ln (2)}$ by $\frac{1}{2}$ .

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

Step 4.3.4.2

Reorder $\ln (2)$ and $2$ .

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

Step 4.3.4.3

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

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

Step 4.3.5

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

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

Step 4.3.6

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

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

Step 4.3.7

Exponentiation and log are inverse functions.

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

Step 4.4

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

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