Trigonometry Examples

$y = \ln (x - 6)$

Step 1

Interchange the variables.

$x = \ln (y - 6)$

Step 2

Solve for $y$ .

Tap for more steps...

Step 2.1

Rewrite the equation as $\ln (y - 6) = x$ .

$\ln (y - 6) = x$

Step 2.2

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

$e^{\ln (y - 6)} = e^{x}$

Step 2.3

Rewrite $\ln (y - 6) = x$ 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 $b^{y} = x$ .

$e^{x} = y - 6$

Step 2.4

Solve for $y$ .

Tap for more steps...

Step 2.4.1

Rewrite the equation as $y - 6 = e^{x}$ .

$y - 6 = e^{x}$

Step 2.4.2

Add $6$ to both sides of the equation.

$y = e^{x} + 6$

Step 3

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

$f^{- 1} (x) = e^{x} + 6$

Step 4

Verify if $f^{- 1} (x) = e^{x} + 6$ is the inverse of $f (x) = \ln (x - 6)$ .

Tap for more steps...

Step 4.1

To verify the inverse, check if $f^{- 1} (f (x)) = x$ and $f (f^{- 1} (x)) = x$ .

Step 4.2

Evaluate $f^{- 1} (f (x))$ .

Tap for more steps...

Step 4.2.1

Set up the composite result function.

$f^{- 1} (f (x))$

Step 4.2.2

Evaluate $f^{- 1} (\ln (x - 6))$ by substituting in the value of $f$ into $f^{- 1}$ .

$f^{- 1} (\ln (x - 6)) = e^{\ln (x - 6)} + 6$

Step 4.2.3

Exponentiation and log are inverse functions.

$f^{- 1} (\ln (x - 6)) = x - 6 + 6$

Step 4.2.4

Combine the opposite terms in $x - 6 + 6$ .

Tap for more steps...

Step 4.2.4.1

Add $- 6$ and $6$ .

$f^{- 1} (\ln (x - 6)) = x + 0$

Step 4.2.4.2

Add $x$ and $0$ .

$f^{- 1} (\ln (x - 6)) = x$

Step 4.3

Evaluate $f (f^{- 1} (x))$ .

Tap for more steps...

Step 4.3.1

Set up the composite result function.

$f (f^{- 1} (x))$

Step 4.3.2

Evaluate $f (e^{x} + 6)$ by substituting in the value of $f^{- 1}$ into $f$ .

$f (e^{x} + 6) = \ln ((e^{x} + 6) - 6)$

Step 4.3.3

Combine the opposite terms in $(e^{x} + 6) - 6$ .

Tap for more steps...

Step 4.3.3.1

Subtract $6$ from $6$ .

$f (e^{x} + 6) = \ln (e^{x} + 0)$

Step 4.3.3.2

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

$f (e^{x} + 6) = \ln (e^{x})$

Step 4.3.4

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

$f (e^{x} + 6) = x \ln (e)$

Step 4.3.5

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

$f (e^{x} + 6) = x \cdot 1$

Step 4.3.6

Multiply $x$ by $1$ .

$f (e^{x} + 6) = x$

Step 4.4

Since $f^{- 1} (f (x)) = x$ and $f (f^{- 1} (x)) = x$ , then $f^{- 1} (x) = e^{x} + 6$ is the inverse of $f (x) = \ln (x - 6)$ .

$f^{- 1} (x) = e^{x} + 6$