Algebra Examples

$y = \frac{1}{3} \cdot 7^{x - 1}$

Step 1

Interchange the variables.

$x = \frac{1}{3} \cdot 7^{y - 1}$

Step 2

Solve for $y$ .

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Step 2.1

Rewrite the equation as $\frac{1}{3} \cdot 7^{y - 1} = x$ .

$\frac{1}{3} \cdot 7^{y - 1} = x$

Step 2.2

Multiply both sides of the equation by $3$ .

$3 (\frac{1}{3} \cdot 7^{y - 1}) = 3 x$

Step 2.3

Simplify the left side.

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Step 2.3.1

Simplify $3 (\frac{1}{3} \cdot 7^{y - 1})$ .

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Step 2.3.1.1

Combine $\frac{1}{3}$ and $7^{y - 1}$ .

$3 \frac{7^{y - 1}}{3} = 3 x$

Step 2.3.1.2

Cancel the common factor of $3$ .

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Step 2.3.1.2.1

Cancel the common factor.

$3 \frac{7^{y - 1}}{3} = 3 x$

Step 2.3.1.2.2

Rewrite the expression.

$7^{y - 1} = 3 x$

Step 2.4

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

$\ln (7^{y - 1}) = \ln (3 x)$

Step 2.5

Expand $\ln (7^{y - 1})$ by moving $y - 1$ outside the logarithm.

$(y - 1) \ln (7) = \ln (3 x)$

Step 2.6

Simplify the left side.

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Step 2.6.1

Simplify $(y - 1) \ln (7)$ .

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Step 2.6.1.1

Apply the distributive property.

$y \ln (7) - 1 \ln (7) = \ln (3 x)$

Step 2.6.1.2

Rewrite $- 1 \ln (7)$ as $- \ln (7)$ .

$y \ln (7) - \ln (7) = \ln (3 x)$

Step 2.7

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

$y \ln (7) - \ln (7) - \ln (3 x) = 0$

Step 2.8

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

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Step 2.8.1

Add $\ln (7)$ to both sides of the equation.

$y \ln (7) - \ln (3 x) = \ln (7)$

Step 2.8.2

Add $\ln (3 x)$ to both sides of the equation.

$y \ln (7) = \ln (7) + \ln (3 x)$

Step 2.9

Divide each term in $y \ln (7) = \ln (7) + \ln (3 x)$ by $\ln (7)$ and simplify.

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Step 2.9.1

Divide each term in $y \ln (7) = \ln (7) + \ln (3 x)$ by $\ln (7)$ .

$\frac{y \ln (7)}{\ln (7)} = \frac{\ln (7)}{\ln (7)} + \frac{\ln (3 x)}{\ln (7)}$

Step 2.9.2

Simplify the left side.

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Step 2.9.2.1

Cancel the common factor of $\ln (7)$ .

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Step 2.9.2.1.1

Cancel the common factor.

$\frac{y \ln (7)}{\ln (7)} = \frac{\ln (7)}{\ln (7)} + \frac{\ln (3 x)}{\ln (7)}$

Step 2.9.2.1.2

Divide $y$ by $1$ .

$y = \frac{\ln (7)}{\ln (7)} + \frac{\ln (3 x)}{\ln (7)}$

Step 2.9.3

Simplify the right side.

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Step 2.9.3.1

Cancel the common factor of $\ln (7)$ .

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Step 2.9.3.1.1

Cancel the common factor.

$y = \frac{\ln (7)}{\ln (7)} + \frac{\ln (3 x)}{\ln (7)}$

Step 2.9.3.1.2

Rewrite the expression.

$y = 1 + \frac{\ln (3 x)}{\ln (7)}$

Step 3

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

$f^{- 1} (x) = 1 + \frac{\ln (3 x)}{\ln (7)}$

Step 4

Verify if $f^{- 1} (x) = 1 + \frac{\ln (3 x)}{\ln (7)}$ is the inverse of $f (x) = \frac{1}{3} \cdot 7^{x - 1}$ .

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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 $f^{- 1} (\frac{1}{3} \cdot 7^{x - 1})$ by substituting in the value of $f$ into $f^{- 1}$ .

$f^{- 1} (\frac{1}{3} \cdot 7^{x - 1}) = 1 + \frac{\ln (3 (\frac{1}{3} \cdot 7^{x - 1}))}{\ln (7)}$

Step 4.2.3

Simplify each term.

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Step 4.2.3.1

Simplify the numerator.

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Step 4.2.3.1.1

Cancel the common factor of $3$ .

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Step 4.2.3.1.1.1

Cancel the common factor.

$f^{- 1} (\frac{1}{3} \cdot 7^{x - 1}) = 1 + \frac{\ln (3 (\frac{1}{3}) \cdot 7^{x - 1})}{\ln (7)}$

Step 4.2.3.1.1.2

Rewrite the expression.

$f^{- 1} (\frac{1}{3} \cdot 7^{x - 1}) = 1 + \frac{\ln (1 \cdot 7^{x - 1})}{\ln (7)}$

Step 4.2.3.1.2

Multiply $7^{x - 1}$ by $1$ .

$f^{- 1} (\frac{1}{3} \cdot 7^{x - 1}) = 1 + \frac{\ln (7^{x - 1})}{\ln (7)}$

Step 4.2.3.2

Expand $\ln (7^{x - 1})$ by moving $x - 1$ outside the logarithm.

$f^{- 1} (\frac{1}{3} \cdot 7^{x - 1}) = 1 + \frac{(x - 1) \ln (7)}{\ln (7)}$

Step 4.2.3.3

Cancel the common factor of $\ln (7)$ .

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Step 4.2.3.3.1

Cancel the common factor.

$f^{- 1} (\frac{1}{3} \cdot 7^{x - 1}) = 1 + \frac{(x - 1) \ln (7)}{\ln (7)}$

Step 4.2.3.3.2

Divide $x - 1$ by $1$ .

$f^{- 1} (\frac{1}{3} \cdot 7^{x - 1}) = 1 + x - 1$

Step 4.2.4

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

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Step 4.2.4.1

Subtract $1$ from $1$ .

$f^{- 1} (\frac{1}{3} \cdot 7^{x - 1}) = x + 0$

Step 4.2.4.2

Add $x$ and $0$ .

$f^{- 1} (\frac{1}{3} \cdot 7^{x - 1}) = x$

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 $f (1 + \frac{\ln (3 x)}{\ln (7)})$ by substituting in the value of $f^{- 1}$ into $f$ .

$f (1 + \frac{\ln (3 x)}{\ln (7)}) = \frac{1}{3} \cdot 7^{(1 + \frac{\ln (3 x)}{\ln (7)}) - 1}$

Step 4.3.3

Combine the opposite terms in $(1 + \frac{\ln (3 x)}{\ln (7)}) - 1$ .

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Step 4.3.3.1

Subtract $1$ from $1$ .

$f (1 + \frac{\ln (3 x)}{\ln (7)}) = \frac{1}{3} \cdot 7^{0 + \frac{\ln (3 x)}{\ln (7)}}$

Step 4.3.3.2

Add $0$ and $\frac{\ln (3 x)}{\ln (7)}$ .

$f (1 + \frac{\ln (3 x)}{\ln (7)}) = \frac{1}{3} \cdot 7^{\frac{\ln (3 x)}{\ln (7)}}$

Step 4.3.4

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

$f (1 + \frac{\ln (3 x)}{\ln (7)}) = \frac{1}{3} \cdot 7^{\log_{7} (3 x)}$

Step 4.3.5

Exponentiation and log are inverse functions.

$f (1 + \frac{\ln (3 x)}{\ln (7)}) = \frac{1}{3} \cdot (3 x)$

Step 4.3.6

Cancel the common factor of $3$ .

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Step 4.3.6.1

Factor $3$ out of $3 x$ .

$f (1 + \frac{\ln (3 x)}{\ln (7)}) = \frac{1}{3} \cdot (3 (x))$

Step 4.3.6.2

Cancel the common factor.

$f (1 + \frac{\ln (3 x)}{\ln (7)}) = \frac{1}{3} \cdot (3 x)$

Step 4.3.6.3

Rewrite the expression.

$f (1 + \frac{\ln (3 x)}{\ln (7)}) = x$

Step 4.4

Since $f^{- 1} (f (x)) = x$ and $f (f^{- 1} (x)) = x$ , then $f^{- 1} (x) = 1 + \frac{\ln (3 x)}{\ln (7)}$ is the inverse of $f (x) = \frac{1}{3} \cdot 7^{x - 1}$ .

$f^{- 1} (x) = 1 + \frac{\ln (3 x)}{\ln (7)}$