Precalculus Examples

$\log_{5} (2 x - 1)$

Step 1

Interchange the variables.

$x = \log_{5} (2 y - 1)$

Step 2

Solve for $y$ .

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Step 2.1

Rewrite the equation as $\log_{5} (2 y - 1) = x$ .

$\log_{5} (2 y - 1) = x$

Step 2.2

Rewrite $\log_{5} (2 y - 1) = x$ in exponential form using the definition of a logarithm. If $x$ and $b$ are positive real numbers and $b \neq 1$ , then $\log_{b} (x) = y$ is equivalent to $b^{y} = x$ .

$5^{x} = 2 y - 1$

Step 2.3

Solve for $y$ .

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Step 2.3.1

Rewrite the equation as $2 y - 1 = 5^{x}$ .

$2 y - 1 = 5^{x}$

Step 2.3.2

Add $1$ to both sides of the equation.

$2 y = 5^{x} + 1$

Step 2.3.3

Divide each term in $2 y = 5^{x} + 1$ by $2$ and simplify.

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Step 2.3.3.1

Divide each term in $2 y = 5^{x} + 1$ by $2$ .

$\frac{2 y}{2} = \frac{5^{x}}{2} + \frac{1}{2}$

Step 2.3.3.2

Simplify the left side.

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Step 2.3.3.2.1

Cancel the common factor of $2$ .

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Step 2.3.3.2.1.1

Cancel the common factor.

$\frac{2 y}{2} = \frac{5^{x}}{2} + \frac{1}{2}$

Step 2.3.3.2.1.2

Divide $y$ by $1$ .

$y = \frac{5^{x}}{2} + \frac{1}{2}$

Step 3

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

$f^{- 1} (x) = \frac{5^{x}}{2} + \frac{1}{2}$

Step 4

Verify if $f^{- 1} (x) = \frac{5^{x}}{2} + \frac{1}{2}$ is the inverse of $f (x) = \log_{5} (2 x - 1)$ .

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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 $f^{- 1} (\log_{5} (2 x - 1))$ by substituting in the value of $f$ into $f^{- 1}$ .

$f^{- 1} (\log_{5} (2 x - 1)) = \frac{5^{\log_{5} (2 x - 1)}}{2} + \frac{1}{2}$

Step 4.2.3

Combine the numerators over the common denominator.

$f^{- 1} (\log_{5} (2 x - 1)) = \frac{5^{\log_{5} (2 x - 1)} + 1}{2}$

Step 4.2.4

Exponentiation and log are inverse functions.

$f^{- 1} (\log_{5} (2 x - 1)) = \frac{2 x - 1 + 1}{2}$

Step 4.2.5

Combine the opposite terms in $2 x - 1 + 1$ .

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Step 4.2.5.1

Add $- 1$ and $1$ .

$f^{- 1} (\log_{5} (2 x - 1)) = \frac{2 x + 0}{2}$

Step 4.2.5.2

Add $2 x$ and $0$ .

$f^{- 1} (\log_{5} (2 x - 1)) = \frac{2 x}{2}$

Step 4.2.6

Cancel the common factor of $2$ .

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Step 4.2.6.1

Cancel the common factor.

$f^{- 1} (\log_{5} (2 x - 1)) = \frac{2 x}{2}$

Step 4.2.6.2

Divide $x$ by $1$ .

$f^{- 1} (\log_{5} (2 x - 1)) = x$

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 $f (\frac{5^{x}}{2} + \frac{1}{2})$ by substituting in the value of $f^{- 1}$ into $f$ .

$f (\frac{5^{x}}{2} + \frac{1}{2}) = \log_{5} (2 (\frac{5^{x}}{2} + \frac{1}{2}) - 1)$

Step 4.3.3

Simplify each term.

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Step 4.3.3.1

Apply the distributive property.

$f (\frac{5^{x}}{2} + \frac{1}{2}) = \log_{5} (2 (\frac{5^{x}}{2}) + 2 (\frac{1}{2}) - 1)$

Step 4.3.3.2

Cancel the common factor of $2$ .

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Step 4.3.3.2.1

Cancel the common factor.

$f (\frac{5^{x}}{2} + \frac{1}{2}) = \log_{5} (2 (\frac{5^{x}}{2}) + 2 (\frac{1}{2}) - 1)$

Step 4.3.3.2.2

Rewrite the expression.

$f (\frac{5^{x}}{2} + \frac{1}{2}) = \log_{5} (5^{x} + 2 (\frac{1}{2}) - 1)$

Step 4.3.3.3

Cancel the common factor of $2$ .

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Step 4.3.3.3.1

Cancel the common factor.

$f (\frac{5^{x}}{2} + \frac{1}{2}) = \log_{5} (5^{x} + 2 (\frac{1}{2}) - 1)$

Step 4.3.3.3.2

Rewrite the expression.

$f (\frac{5^{x}}{2} + \frac{1}{2}) = \log_{5} (5^{x} + 1 - 1)$

Step 4.3.4

Combine the opposite terms in $5^{x} + 1 - 1$ .

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Step 4.3.4.1

Subtract $1$ from $1$ .

$f (\frac{5^{x}}{2} + \frac{1}{2}) = \log_{5} (5^{x} + 0)$

Step 4.3.4.2

Add $5^{x}$ and $0$ .

$f (\frac{5^{x}}{2} + \frac{1}{2}) = \log_{5} (5^{x})$

Step 4.3.5

Use logarithm rules to move $x$ out of the exponent.

$f (\frac{5^{x}}{2} + \frac{1}{2}) = x \log_{5} (5)$

Step 4.3.6

Logarithm base $5$ of $5$ is $1$ .

$f (\frac{5^{x}}{2} + \frac{1}{2}) = x \cdot 1$

Step 4.3.7

Multiply $x$ by $1$ .

$f (\frac{5^{x}}{2} + \frac{1}{2}) = x$

Step 4.4

Since $f^{- 1} (f (x)) = x$ and $f (f^{- 1} (x)) = x$ , then $f^{- 1} (x) = \frac{5^{x}}{2} + \frac{1}{2}$ is the inverse of $f (x) = \log_{5} (2 x - 1)$ .

$f^{- 1} (x) = \frac{5^{x}}{2} + \frac{1}{2}$