Trigonometry Examples

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

Step 1

Write $f (x) = \log_{5} (x + 1)$ as an equation.

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

Step 2

Interchange the variables.

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

Step 3

Solve for $y$ .

Tap for more steps...

Step 3.1

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

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

Step 3.2

Rewrite $\log_{5} (y + 1) = x$ 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 $b^{y} = x$ .

$5^{x} = y + 1$

Step 3.3

Solve for $y$ .

Tap for more steps...

Step 3.3.1

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

$y + 1 = 5^{x}$

Step 3.3.2

Subtract $1$ from both sides of the equation.

$y = 5^{x} - 1$

Step 4

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

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

Step 5

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

Tap for more steps...

Step 5.1

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

Step 5.2

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

Tap for more steps...

Step 5.2.1

Set up the composite result function.

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

Step 5.2.2

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

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

Step 5.2.3

Exponentiation and log are inverse functions.

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

Step 5.2.4

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

Tap for more steps...

Step 5.2.4.1

Subtract $1$ from $1$ .

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

Step 5.2.4.2

Add $x$ and $0$ .

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

Step 5.3

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

Tap for more steps...

Step 5.3.1

Set up the composite result function.

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

Step 5.3.2

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

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

Step 5.3.3

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

Tap for more steps...

Step 5.3.3.1

Add $- 1$ and $1$ .

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

Step 5.3.3.2

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

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

Step 5.3.4

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

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

Step 5.3.5

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

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

Step 5.3.6

Multiply $x$ by $1$ .

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

Step 5.4

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

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