Trigonometry Examples

$y = \log_{2} (2 x - 3)$

Step 1

Interchange the variables.

$x = \log_{2} (2 y - 3)$

Step 2

Solve for $y$ .

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Step 2.1

Rewrite the equation as $\log_{2} (2 y - 3) = x$ .

$\log_{2} (2 y - 3) = x$

Step 2.2

Rewrite $\log_{2} (2 y - 3) = x$ 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 $b^{y} = x$ .

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

Step 2.3

Solve for $y$ .

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Step 2.3.1

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

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

Step 2.3.2

Add $3$ to both sides of the equation.

$2 y = 2^{x} + 3$

Step 2.3.3

Divide each term in $2 y = 2^{x} + 3$ by $2$ and simplify.

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Step 2.3.3.1

Divide each term in $2 y = 2^{x} + 3$ by $2$ .

$\frac{2 y}{2} = \frac{2^{x}}{2} + \frac{3}{2}$

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.

$\frac{2 y}{2} = \frac{2^{x}}{2} + \frac{3}{2}$

Step 2.3.3.2.1.2

Divide $y$ by $1$ .

$y = \frac{2^{x}}{2} + \frac{3}{2}$

Step 2.3.3.3

Simplify the right side.

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Step 2.3.3.3.1

Cancel the common factor of $2^{x}$ and $2$ .

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Step 2.3.3.3.1.1

Factor $2$ out of $2^{x}$ .

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

Step 2.3.3.3.1.2

Cancel the common factors.

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Step 2.3.3.3.1.2.1

Factor $2$ out of $2$ .

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

Step 2.3.3.3.1.2.2

Cancel the common factor.

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

Step 2.3.3.3.1.2.3

Rewrite the expression.

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

Step 2.3.3.3.1.2.4

Divide $2^{x - 1}$ by $1$ .

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

Step 2.3.3.3.2

To write $2^{x - 1}$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

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

Step 2.3.3.3.3

Combine $2^{x - 1}$ and $\frac{2}{2}$ .

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

Step 2.3.3.3.4

Combine the numerators over the common denominator.

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

Step 2.3.3.3.5

Multiply $2^{x - 1}$ by $2$ by adding the exponents.

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Step 2.3.3.3.5.1

Multiply $2^{x - 1}$ by $2$ .

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Step 2.3.3.3.5.1.1

Raise $2$ to the power of $1$ .

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

Step 2.3.3.3.5.1.2

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

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

Step 2.3.3.3.5.2

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

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Step 2.3.3.3.5.2.1

Add $- 1$ and $1$ .

$y = \frac{2^{x + 0} + 3}{2}$

Step 2.3.3.3.5.2.2

Add $x$ and $0$ .

$y = \frac{2^{x} + 3}{2}$

Step 3

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

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

Step 4

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

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

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

Step 4.2.3

Simplify the numerator.

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Step 4.2.3.1

Exponentiation and log are inverse functions.

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

Step 4.2.3.2

Add $- 3$ and $3$ .

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

Step 4.2.3.3

Add $2 x$ and $0$ .

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

Step 4.2.4

Cancel the common factor of $2$ .

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Step 4.2.4.1

Cancel the common factor.

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

Step 4.2.4.2

Divide $x$ by $1$ .

$f^{- 1} (\log_{2} (2 x - 3)) = x$

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

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

Step 4.3.3

Cancel the common factor of $2$ .

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Step 4.3.3.1

Cancel the common factor.

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

Step 4.3.3.2

Rewrite the expression.

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

Step 4.3.4

Combine the opposite terms in $2^{x} + 3 - 3$ .

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Step 4.3.4.1

Subtract $3$ from $3$ .

$f (\frac{2^{x} + 3}{2}) = \log_{2} (2^{x} + 0)$

Step 4.3.4.2

Add $2^{x}$ and $0$ .

$f (\frac{2^{x} + 3}{2}) = \log_{2} (2^{x})$

Step 4.3.5

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

$f (\frac{2^{x} + 3}{2}) = x \log_{2} (2)$

Step 4.3.6

Logarithm base $2$ of $2$ is $1$ .

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

Step 4.3.7

Multiply $x$ by $1$ .

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

Step 4.4

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

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