Algebra Examples

$y = 4^{2 x + 9}$

Step 1

Interchange the variables.

$x = 4^{2 y + 9}$

Step 2

Solve for $y$ .

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Step 2.1

Rewrite the equation as $4^{2 y + 9} = x$ .

$4^{2 y + 9} = x$

Step 2.2

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

$\ln (4^{2 y + 9}) = \ln (x)$

Step 2.3

Expand $\ln (4^{2 y + 9})$ by moving $2 y + 9$ outside the logarithm.

$(2 y + 9) \ln (4) = \ln (x)$

Step 2.4

Simplify the left side.

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Step 2.4.1

Apply the distributive property.

$2 y \ln (4) + 9 \ln (4) = \ln (x)$

Step 2.5

Move all the terms containing a logarithm to the left side of the equation.

$2 y \ln (4) + 9 \ln (4) - \ln (x) = 0$

Step 2.6

Move all terms not containing $y$ to the right side of the equation.

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Step 2.6.1

Subtract $9 \ln (4)$ from both sides of the equation.

$2 y \ln (4) - \ln (x) = - 9 \ln (4)$

Step 2.6.2

Add $\ln (x)$ to both sides of the equation.

$2 y \ln (4) = - 9 \ln (4) + \ln (x)$

Step 2.7

Divide each term in $2 y \ln (4) = - 9 \ln (4) + \ln (x)$ by $2 \ln (4)$ and simplify.

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Step 2.7.1

Divide each term in $2 y \ln (4) = - 9 \ln (4) + \ln (x)$ by $2 \ln (4)$ .

$\frac{2 y \ln (4)}{2 \ln (4)} = \frac{- 9 \ln (4)}{2 \ln (4)} + \frac{\ln (x)}{2 \ln (4)}$

Step 2.7.2

Simplify the left side.

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Step 2.7.2.1

Cancel the common factor of $2$ .

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Step 2.7.2.1.1

Cancel the common factor.

$\frac{2 y \ln (4)}{2 \ln (4)} = \frac{- 9 \ln (4)}{2 \ln (4)} + \frac{\ln (x)}{2 \ln (4)}$

Step 2.7.2.1.2

Rewrite the expression.

$\frac{y \ln (4)}{\ln (4)} = \frac{- 9 \ln (4)}{2 \ln (4)} + \frac{\ln (x)}{2 \ln (4)}$

Step 2.7.2.2

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

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Step 2.7.2.2.1

Cancel the common factor.

$\frac{y \ln (4)}{\ln (4)} = \frac{- 9 \ln (4)}{2 \ln (4)} + \frac{\ln (x)}{2 \ln (4)}$

Step 2.7.2.2.2

Divide $y$ by $1$ .

$y = \frac{- 9 \ln (4)}{2 \ln (4)} + \frac{\ln (x)}{2 \ln (4)}$

Step 2.7.3

Simplify the right side.

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Step 2.7.3.1

Simplify each term.

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Step 2.7.3.1.1

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

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Step 2.7.3.1.1.1

Cancel the common factor.

$y = \frac{- 9 \ln (4)}{2 \ln (4)} + \frac{\ln (x)}{2 \ln (4)}$

Step 2.7.3.1.1.2

Rewrite the expression.

$y = \frac{- 9}{2} + \frac{\ln (x)}{2 \ln (4)}$

Step 2.7.3.1.2

Move the negative in front of the fraction.

$y = - \frac{9}{2} + \frac{\ln (x)}{2 \ln (4)}$

Step 3

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

$f^{- 1} (x) = - \frac{9}{2} + \frac{\ln (x)}{2 \ln (4)}$

Step 4

Verify if $f^{- 1} (x) = - \frac{9}{2} + \frac{\ln (x)}{2 \ln (4)}$ is the inverse of $f (x) = 4^{2 x + 9}$ .

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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 $f^{- 1} \cdot 4^{2 x + 9}$ by substituting in the value of $f$ into $f^{- 1}$ .

$f^{- 1} \cdot 4^{2 x + 9} = - \frac{9}{2} + \frac{\ln (4^{2 x + 9})}{2 \ln (4)}$

Step 4.2.3

Simplify each term.

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Step 4.2.3.1

Expand $\ln (4^{2 x + 9})$ by moving $2 x + 9$ outside the logarithm.

$f^{- 1} \cdot 4^{2 x + 9} = - \frac{9}{2} + \frac{(2 x + 9) \ln (4)}{2 \ln (4)}$

Step 4.2.3.2

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

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Step 4.2.3.2.1

Cancel the common factor.

$f^{- 1} \cdot 4^{2 x + 9} = - \frac{9}{2} + \frac{(2 x + 9) \ln (4)}{2 \ln (4)}$

Step 4.2.3.2.2

Rewrite the expression.

$f^{- 1} \cdot 4^{2 x + 9} = - \frac{9}{2} + \frac{2 x + 9}{2}$

Step 4.2.4

Simplify terms.

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Step 4.2.4.1

Combine the numerators over the common denominator.

$f^{- 1} \cdot 4^{2 x + 9} = \frac{- 9 + 2 x + 9}{2}$

Step 4.2.4.2

Combine the opposite terms in $- 9 + 2 x + 9$ .

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Step 4.2.4.2.1

Add $- 9$ and $9$ .

$f^{- 1} \cdot 4^{2 x + 9} = \frac{2 x + 0}{2}$

Step 4.2.4.2.2

Add $2 x$ and $0$ .

$f^{- 1} \cdot 4^{2 x + 9} = \frac{2 x}{2}$

Step 4.2.4.3

Cancel the common factor of $2$ .

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Step 4.2.4.3.1

Cancel the common factor.

$f^{- 1} \cdot 4^{2 x + 9} = \frac{2 x}{2}$

Step 4.2.4.3.2

Divide $x$ by $1$ .

$f^{- 1} \cdot 4^{2 x + 9} = x$

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 $f (- \frac{9}{2} + \frac{\ln (x)}{2 \ln (4)})$ by substituting in the value of $f^{- 1}$ into $f$ .

$f (- \frac{9}{2} + \frac{\ln (x)}{2 \ln (4)}) = 4^{2 (- \frac{9}{2} + \frac{\ln (x)}{2 \ln (4)}) + 9}$

Step 4.3.3

Simplify each term.

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Step 4.3.3.1

Simplify each term.

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Step 4.3.3.1.1

Simplify $2 \ln (4)$ by moving $2$ inside the logarithm.

$f (- \frac{9}{2} + \frac{\ln (x)}{2 \ln (4)}) = 4^{2 (- \frac{9}{2} + \frac{\ln (x)}{\ln (4^{2})}) + 9}$

Step 4.3.3.1.2

Raise $4$ to the power of $2$ .

$f (- \frac{9}{2} + \frac{\ln (x)}{2 \ln (4)}) = 4^{2 (- \frac{9}{2} + \frac{\ln (x)}{\ln (16)}) + 9}$

Step 4.3.3.2

Apply the distributive property.

$f (- \frac{9}{2} + \frac{\ln (x)}{2 \ln (4)}) = 4^{2 (- \frac{9}{2}) + 2 (\frac{\ln (x)}{\ln (16)}) + 9}$

Step 4.3.3.3

Cancel the common factor of $2$ .

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Step 4.3.3.3.1

Move the leading negative in $- \frac{9}{2}$ into the numerator.

$f (- \frac{9}{2} + \frac{\ln (x)}{2 \ln (4)}) = 4^{2 (\frac{- 9}{2}) + 2 (\frac{\ln (x)}{\ln (16)}) + 9}$

Step 4.3.3.3.2

Cancel the common factor.

$f (- \frac{9}{2} + \frac{\ln (x)}{2 \ln (4)}) = 4^{2 (\frac{- 9}{2}) + 2 (\frac{\ln (x)}{\ln (16)}) + 9}$

Step 4.3.3.3.3

Rewrite the expression.

$f (- \frac{9}{2} + \frac{\ln (x)}{2 \ln (4)}) = 4^{- 9 + 2 (\frac{\ln (x)}{\ln (16)}) + 9}$

Step 4.3.3.4

Multiply $2 \frac{\ln (x)}{\ln (16)}$ .

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Step 4.3.3.4.1

Combine $2$ and $\frac{\ln (x)}{\ln (16)}$ .

$f (- \frac{9}{2} + \frac{\ln (x)}{2 \ln (4)}) = 4^{- 9 + \frac{2 \ln (x)}{\ln (16)} + 9}$

Step 4.3.3.4.2

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

$f (- \frac{9}{2} + \frac{\ln (x)}{2 \ln (4)}) = 4^{- 9 + \frac{\ln (x^{2})}{\ln (16)} + 9}$

Step 4.3.4

Combine the opposite terms in $- 9 + \frac{\ln (x^{2})}{\ln (16)} + 9$ .

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Step 4.3.4.1

Add $- 9$ and $9$ .

$f (- \frac{9}{2} + \frac{\ln (x)}{2 \ln (4)}) = 4^{0 + \frac{\ln (x^{2})}{\ln (16)}}$

Step 4.3.4.2

Add $0$ and $\frac{\ln (x^{2})}{\ln (16)}$ .

$f (- \frac{9}{2} + \frac{\ln (x)}{2 \ln (4)}) = 4^{\frac{\ln (x^{2})}{\ln (16)}}$

Step 4.4

Since $f^{- 1} (f (x)) = x$ and $f (f^{- 1} (x)) = x$ , then $f^{- 1} (x) = - \frac{9}{2} + \frac{\ln (x)}{2 \ln (4)}$ is the inverse of $f (x) = 4^{2 x + 9}$ .

$f^{- 1} (x) = - \frac{9}{2} + \frac{\ln (x)}{2 \ln (4)}$