Algebra Examples

$y = \log_{2} (x + 7)$

Step 1

Interchange the variables.

$x = \log_{2} (y + 7)$

Step 2

Solve for $y$ .

Tap for more steps...

Step 2.1

Rewrite the equation as $\log_{2} (y + 7) = x$ .

$\log_{2} (y + 7) = x$

Step 2.2

Rewrite $\log_{2} (y + 7) = x$ 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 $b^{y} = x$ .

$2^{x} = y + 7$

Step 2.3

Solve for $y$ .

Tap for more steps...

Step 2.3.1

Rewrite the equation as $y + 7 = 2^{x}$ .

$y + 7 = 2^{x}$

Step 2.3.2

Subtract $7$ from both sides of the equation.

$y = 2^{x} - 7$

Step 3

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

$f^{- 1} (x) = 2^{x} - 7$

Step 4

Verify if $f^{- 1} (x) = 2^{x} - 7$ is the inverse of $f (x) = \log_{2} (x + 7)$ .

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 $f^{- 1} (\log_{2} (x + 7))$ by substituting in the value of $f$ into $f^{- 1}$ .

$f^{- 1} (\log_{2} (x + 7)) = 2^{\log_{2} (x + 7)} - 7$

Step 4.2.3

Exponentiation and log are inverse functions.

$f^{- 1} (\log_{2} (x + 7)) = x + 7 - 7$

Step 4.2.4

Combine the opposite terms in $x + 7 - 7$ .

Tap for more steps...

Step 4.2.4.1

Subtract $7$ from $7$ .

$f^{- 1} (\log_{2} (x + 7)) = x + 0$

Step 4.2.4.2

Add $x$ and $0$ .

$f^{- 1} (\log_{2} (x + 7)) = x$

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 $f (2^{x} - 7)$ by substituting in the value of $f^{- 1}$ into $f$ .

$f (2^{x} - 7) = \log_{2} ((2^{x} - 7) + 7)$

Step 4.3.3

Combine the opposite terms in $(2^{x} - 7) + 7$ .

Tap for more steps...

Step 4.3.3.1

Add $- 7$ and $7$ .

$f (2^{x} - 7) = \log_{2} (2^{x} + 0)$

Step 4.3.3.2

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

$f (2^{x} - 7) = \log_{2} (2^{x})$

Step 4.3.4

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

$f (2^{x} - 7) = x \log_{2} (2)$

Step 4.3.5

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

$f (2^{x} - 7) = x \cdot 1$

Step 4.3.6

Multiply $x$ by $1$ .

$f (2^{x} - 7) = x$

Step 4.4

Since $f^{- 1} (f (x)) = x$ and $f (f^{- 1} (x)) = x$ , then $f^{- 1} (x) = 2^{x} - 7$ is the inverse of $f (x) = \log_{2} (x + 7)$ .

$f^{- 1} (x) = 2^{x} - 7$