Trigonometry Examples

$y = 2 \ln (x - 5)$

Step 1

Interchange the variables.

$x = 2 \ln (y - 5)$

Step 2

Solve for $y$ .

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Step 2.1

Rewrite the equation as $2 \ln (y - 5) = x$ .

$2 \ln (y - 5) = x$

Step 2.2

Divide each term in $2 \ln (y - 5) = x$ by $2$ and simplify.

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Step 2.2.1

Divide each term in $2 \ln (y - 5) = x$ by $2$ .

$\frac{2 \ln (y - 5)}{2} = \frac{x}{2}$

Step 2.2.2

Simplify the left side.

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Step 2.2.2.1

Cancel the common factor of $2$ .

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Step 2.2.2.1.1

Cancel the common factor.

$\frac{2 \ln (y - 5)}{2} = \frac{x}{2}$

Step 2.2.2.1.2

Divide $\ln (y - 5)$ by $1$ .

$\ln (y - 5) = \frac{x}{2}$

Step 2.3

To solve for $y$ , rewrite the equation using properties of logarithms.

$e^{\ln (y - 5)} = e^{\frac{x}{2}}$

Step 2.4

Rewrite $\ln (y - 5) = \frac{x}{2}$ 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 $b^{y} = x$ .

$e^{\frac{x}{2}} = y - 5$

Step 2.5

Solve for $y$ .

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Step 2.5.1

Rewrite the equation as $y - 5 = e^{\frac{x}{2}}$ .

$y - 5 = e^{\frac{x}{2}}$

Step 2.5.2

Add $5$ to both sides of the equation.

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

Step 3

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

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

Step 4

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

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

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

Step 4.2.3

Simplify each term.

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Step 4.2.3.1

Cancel the common factor of $2$ .

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Step 4.2.3.1.1

Cancel the common factor.

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

Step 4.2.3.1.2

Divide $\ln (x - 5)$ by $1$ .

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

Step 4.2.3.2

Exponentiation and log are inverse functions.

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

Step 4.2.4

Combine the opposite terms in $x - 5 + 5$ .

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Step 4.2.4.1

Add $- 5$ and $5$ .

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

Step 4.2.4.2

Add $x$ and $0$ .

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

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

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

Step 4.3.3

Combine the opposite terms in $(e^{\frac{x}{2}} + 5) - 5$ .

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Step 4.3.3.1

Subtract $5$ from $5$ .

$f (e^{\frac{x}{2}} + 5) = 2 \ln (e^{\frac{x}{2}} + 0)$

Step 4.3.3.2

Add $e^{\frac{x}{2}}$ and $0$ .

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

Step 4.3.4

Use logarithm rules to move $\frac{x}{2}$ out of the exponent.

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

Step 4.3.5

The natural logarithm of $e$ is $1$ .

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

Step 4.3.6

Multiply $\frac{x}{2}$ by $1$ .

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

Step 4.3.7

Cancel the common factor of $2$ .

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Step 4.3.7.1

Cancel the common factor.

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

Step 4.3.7.2

Rewrite the expression.

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

Step 4.4

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

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