Trigonometry Examples

$y = \log (x - 2)$

Step 1

Interchange the variables.

$x = \log (y - 2)$

Step 2

Solve for $y$ .

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Step 2.1

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

$\log (y - 2) = x$

Step 2.2

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

$10^{x} = y - 2$

Step 2.3

Solve for $y$ .

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Step 2.3.1

Rewrite the equation as $y - 2 = 10^{x}$ .

$y - 2 = 10^{x}$

Step 2.3.2

Add $2$ to both sides of the equation.

$y = 10^{x} + 2$

Step 3

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

$f^{- 1} (x) = 10^{x} + 2$

Step 4

Verify if $f^{- 1} (x) = 10^{x} + 2$ is the inverse of $f (x) = \log (x - 2)$ .

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

$f^{- 1} (\log (x - 2)) = 10^{\log (x - 2)} + 2$

Step 4.2.3

Exponentiation and log are inverse functions.

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

Step 4.2.4

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

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Step 4.2.4.1

Add $- 2$ and $2$ .

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

Step 4.2.4.2

Add $x$ and $0$ .

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

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 $f (10^{x} + 2)$ by substituting in the value of $f^{- 1}$ into $f$ .

$f (10^{x} + 2) = \log ((10^{x} + 2) - 2)$

Step 4.3.3

Combine the opposite terms in $(10^{x} + 2) - 2$ .

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Step 4.3.3.1

Subtract $2$ from $2$ .

$f (10^{x} + 2) = \log (10^{x} + 0)$

Step 4.3.3.2

Add $10^{x}$ and $0$ .

$f (10^{x} + 2) = \log (10^{x})$

Step 4.3.4

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

$f (10^{x} + 2) = x \log (10)$

Step 4.3.5

Logarithm base $10$ of $10$ is $1$ .

$f (10^{x} + 2) = x \cdot 1$

Step 4.3.6

Multiply $x$ by $1$ .

$f (10^{x} + 2) = x$

Step 4.4

Since $f^{- 1} (f (x)) = x$ and $f (f^{- 1} (x)) = x$ , then $f^{- 1} (x) = 10^{x} + 2$ is the inverse of $f (x) = \log (x - 2)$ .

$f^{- 1} (x) = 10^{x} + 2$