Algebra Examples

$y = 2^{3 x - 1}$

Step 1

Interchange the variables.

$x = 2^{3 y - 1}$

Step 2

Solve for $y$ .

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Step 2.1

Rewrite the equation as $2^{3 y - 1} = x$ .

$2^{3 y - 1} = x$

Step 2.2

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

$\ln (2^{3 y - 1}) = \ln (x)$

Step 2.3

Expand $\ln (2^{3 y - 1})$ by moving $3 y - 1$ outside the logarithm.

$(3 y - 1) \ln (2) = \ln (x)$

Step 2.4

Simplify the left side.

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Step 2.4.1

Simplify $(3 y - 1) \ln (2)$ .

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Step 2.4.1.1

Apply the distributive property.

$3 y \ln (2) - 1 \ln (2) = \ln (x)$

Step 2.4.1.2

Rewrite $- 1 \ln (2)$ as $- \ln (2)$ .

$3 y \ln (2) - \ln (2) = \ln (x)$

Step 2.5

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

$3 y \ln (2) - \ln (2) - \ln (x) = 0$

Step 2.6

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

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Step 2.6.1

Add $\ln (2)$ to both sides of the equation.

$3 y \ln (2) - \ln (x) = \ln (2)$

Step 2.6.2

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

$3 y \ln (2) = \ln (2) + \ln (x)$

Step 2.7

Divide each term in $3 y \ln (2) = \ln (2) + \ln (x)$ by $3 \ln (2)$ and simplify.

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Step 2.7.1

Divide each term in $3 y \ln (2) = \ln (2) + \ln (x)$ by $3 \ln (2)$ .

$\frac{3 y \ln (2)}{3 \ln (2)} = \frac{\ln (2)}{3 \ln (2)} + \frac{\ln (x)}{3 \ln (2)}$

Step 2.7.2

Simplify the left side.

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Step 2.7.2.1

Cancel the common factor of $3$ .

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Step 2.7.2.1.1

Cancel the common factor.

$\frac{3 y \ln (2)}{3 \ln (2)} = \frac{\ln (2)}{3 \ln (2)} + \frac{\ln (x)}{3 \ln (2)}$

Step 2.7.2.1.2

Rewrite the expression.

$\frac{y \ln (2)}{\ln (2)} = \frac{\ln (2)}{3 \ln (2)} + \frac{\ln (x)}{3 \ln (2)}$

Step 2.7.2.2

Cancel the common factor of $\ln (2)$ .

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Step 2.7.2.2.1

Cancel the common factor.

$\frac{y \ln (2)}{\ln (2)} = \frac{\ln (2)}{3 \ln (2)} + \frac{\ln (x)}{3 \ln (2)}$

Step 2.7.2.2.2

Divide $y$ by $1$ .

$y = \frac{\ln (2)}{3 \ln (2)} + \frac{\ln (x)}{3 \ln (2)}$

Step 2.7.3

Simplify the right side.

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Step 2.7.3.1

Cancel the common factor of $\ln (2)$ .

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Step 2.7.3.1.1

Cancel the common factor.

$y = \frac{\ln (2)}{3 \ln (2)} + \frac{\ln (x)}{3 \ln (2)}$

Step 2.7.3.1.2

Rewrite the expression.

$y = \frac{1}{3} + \frac{\ln (x)}{3 \ln (2)}$

Step 3

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

$f^{- 1} (x) = \frac{1}{3} + \frac{\ln (x)}{3 \ln (2)}$

Step 4

Verify if $f^{- 1} (x) = \frac{1}{3} + \frac{\ln (x)}{3 \ln (2)}$ is the inverse of $f (x) = 2^{3 x - 1}$ .

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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 $f^{- 1} \cdot 2^{3 x - 1}$ by substituting in the value of $f$ into $f^{- 1}$ .

$f^{- 1} \cdot 2^{3 x - 1} = \frac{1}{3} + \frac{\ln (2^{3 x - 1})}{3 \ln (2)}$

Step 4.2.3

Simplify each term.

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Step 4.2.3.1

Expand $\ln (2^{3 x - 1})$ by moving $3 x - 1$ outside the logarithm.

$f^{- 1} \cdot 2^{3 x - 1} = \frac{1}{3} + \frac{(3 x - 1) \ln (2)}{3 \ln (2)}$

Step 4.2.3.2

Cancel the common factor of $\ln (2)$ .

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Step 4.2.3.2.1

Cancel the common factor.

$f^{- 1} \cdot 2^{3 x - 1} = \frac{1}{3} + \frac{(3 x - 1) \ln (2)}{3 \ln (2)}$

Step 4.2.3.2.2

Rewrite the expression.

$f^{- 1} \cdot 2^{3 x - 1} = \frac{1}{3} + \frac{3 x - 1}{3}$

Step 4.2.4

Simplify terms.

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Step 4.2.4.1

Combine the numerators over the common denominator.

$f^{- 1} \cdot 2^{3 x - 1} = \frac{1 + 3 x - 1}{3}$

Step 4.2.4.2

Combine the opposite terms in $1 + 3 x - 1$ .

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Step 4.2.4.2.1

Subtract $1$ from $1$ .

$f^{- 1} \cdot 2^{3 x - 1} = \frac{3 x + 0}{3}$

Step 4.2.4.2.2

Add $3 x$ and $0$ .

$f^{- 1} \cdot 2^{3 x - 1} = \frac{3 x}{3}$

Step 4.2.4.3

Cancel the common factor of $3$ .

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Step 4.2.4.3.1

Cancel the common factor.

$f^{- 1} \cdot 2^{3 x - 1} = \frac{3 x}{3}$

Step 4.2.4.3.2

Divide $x$ by $1$ .

$f^{- 1} \cdot 2^{3 x - 1} = x$

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 $f (\frac{1}{3} + \frac{\ln (x)}{3 \ln (2)})$ by substituting in the value of $f^{- 1}$ into $f$ .

$f (\frac{1}{3} + \frac{\ln (x)}{3 \ln (2)}) = 2^{3 (\frac{1}{3} + \frac{\ln (x)}{3 \ln (2)}) - 1}$

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 $3 \ln (2)$ by moving $3$ inside the logarithm.

$f (\frac{1}{3} + \frac{\ln (x)}{3 \ln (2)}) = 2^{3 (\frac{1}{3} + \frac{\ln (x)}{\ln (2^{3})}) - 1}$

Step 4.3.3.1.2

Raise $2$ to the power of $3$ .

$f (\frac{1}{3} + \frac{\ln (x)}{3 \ln (2)}) = 2^{3 (\frac{1}{3} + \frac{\ln (x)}{\ln (8)}) - 1}$

Step 4.3.3.2

Apply the distributive property.

$f (\frac{1}{3} + \frac{\ln (x)}{3 \ln (2)}) = 2^{3 (\frac{1}{3}) + 3 (\frac{\ln (x)}{\ln (8)}) - 1}$

Step 4.3.3.3

Cancel the common factor of $3$ .

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Step 4.3.3.3.1

Cancel the common factor.

$f (\frac{1}{3} + \frac{\ln (x)}{3 \ln (2)}) = 2^{3 (\frac{1}{3}) + 3 (\frac{\ln (x)}{\ln (8)}) - 1}$

Step 4.3.3.3.2

Rewrite the expression.

$f (\frac{1}{3} + \frac{\ln (x)}{3 \ln (2)}) = 2^{1 + 3 (\frac{\ln (x)}{\ln (8)}) - 1}$

Step 4.3.3.4

Multiply $3 \frac{\ln (x)}{\ln (8)}$ .

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Step 4.3.3.4.1

Combine $3$ and $\frac{\ln (x)}{\ln (8)}$ .

$f (\frac{1}{3} + \frac{\ln (x)}{3 \ln (2)}) = 2^{1 + \frac{3 \ln (x)}{\ln (8)} - 1}$

Step 4.3.3.4.2

Simplify $3 \ln (x)$ by moving $3$ inside the logarithm.

$f (\frac{1}{3} + \frac{\ln (x)}{3 \ln (2)}) = 2^{1 + \frac{\ln (x^{3})}{\ln (8)} - 1}$

Step 4.3.4

Combine the opposite terms in $1 + \frac{\ln (x^{3})}{\ln (8)} - 1$ .

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Step 4.3.4.1

Subtract $1$ from $1$ .

$f (\frac{1}{3} + \frac{\ln (x)}{3 \ln (2)}) = 2^{0 + \frac{\ln (x^{3})}{\ln (8)}}$

Step 4.3.4.2

Add $0$ and $\frac{\ln (x^{3})}{\ln (8)}$ .

$f (\frac{1}{3} + \frac{\ln (x)}{3 \ln (2)}) = 2^{\frac{\ln (x^{3})}{\ln (8)}}$

Step 4.4

Since $f^{- 1} (f (x)) = x$ and $f (f^{- 1} (x)) = x$ , then $f^{- 1} (x) = \frac{1}{3} + \frac{\ln (x)}{3 \ln (2)}$ is the inverse of $f (x) = 2^{3 x - 1}$ .

$f^{- 1} (x) = \frac{1}{3} + \frac{\ln (x)}{3 \ln (2)}$