Find the Inverse y = log base 2 of x+5-9

$y = \log_{2} (x + 5) - 9$

Interchange the variables.

$x = \log_{2} (y + 5) - 9$

Solve for $y$ .

Rewrite the equation as $\log_{2} (y + 5) - 9 = x$ .

$\log_{2} (y + 5) - 9 = x$

Add $9$ to both sides of the equation.

$\log_{2} (y + 5) = x + 9$

Rewrite $\log_{2} (y + 5) = x + 9$ 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 $b^{y} = x$ .

$2^{x + 9} = y + 5$

Solve for $y$ .

Rewrite the equation as $y + 5 = 2^{x + 9}$ .

$y + 5 = 2^{x + 9}$

Subtract $5$ from both sides of the equation.

$y = 2^{x + 9} - 5$

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

$f^{- 1} (x) = 2^{x + 9} - 5$

Verify if $f^{- 1} (x) = 2^{x + 9} - 5$ is the inverse of $f (x) = \log_{2} (x + 5) - 9$ .

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

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

Set up the composite result function.

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

Evaluate $f^{- 1} (\log_{2} (x + 5) - 9)$ by substituting in the value of $f$ into $f^{- 1}$ .

$f^{- 1} (\log_{2} (x + 5) - 9) = 2^{(\log_{2} (x + 5) - 9) + 9} - 5$

Combine the opposite terms in $2^{(\log_{2} (x + 5) - 9) + 9} - 5$ .

Add $- 9$ and $9$ .

$f^{- 1} (\log_{2} (x + 5) - 9) = 2^{\log_{2} (x + 5) + 0} - 5$

Add $\log_{2} (x + 5)$ and $0$ .

$f^{- 1} (\log_{2} (x + 5) - 9) = 2^{\log_{2} (x + 5)} - 5$

Exponentiation and log are inverse functions.

$f^{- 1} (\log_{2} (x + 5) - 9) = x + 5 - 5$

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

Subtract $5$ from $5$ .

$f^{- 1} (\log_{2} (x + 5) - 9) = x + 0$

Add $x$ and $0$ .

$f^{- 1} (\log_{2} (x + 5) - 9) = x$

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

Set up the composite result function.

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

Evaluate $f (2^{x + 9} - 5)$ by substituting in the value of $f^{- 1}$ into $f$ .

$f (2^{x + 9} - 5) = \log_{2} ((2^{x + 9} - 5) + 5) - 9$

Combine the opposite terms in $\log_{2} ((2^{x + 9} - 5) + 5) - 9$ .

Add $- 5$ and $5$ .

$f (2^{x + 9} - 5) = \log_{2} (2^{x + 9} + 0) - 9$

Add $2^{x + 9}$ and $0$ .

$f (2^{x + 9} - 5) = \log_{2} (2^{x + 9}) - 9$

Simplify each term.

Use logarithm rules to move $x + 9$ out of the exponent.

$f (2^{x + 9} - 5) = (x + 9) \log_{2} (2) - 9$

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

$f (2^{x + 9} - 5) = (x + 9) \cdot 1 - 9$

Multiply $x + 9$ by $1$ .

$f (2^{x + 9} - 5) = x + 9 - 9$

Combine the opposite terms in $x + 9 - 9$ .

Subtract $9$ from $9$ .

$f (2^{x + 9} - 5) = x + 0$

Add $x$ and $0$ .

$f (2^{x + 9} - 5) = x$

Since $f^{- 1} (f (x)) = x$ and $f (f^{- 1} (x)) = x$ , then $f^{- 1} (x) = 2^{x + 9} - 5$ is the inverse of $f (x) = \log_{2} (x + 5) - 9$ .

$f^{- 1} (x) = 2^{x + 9} - 5$

Find the Inverse y = log base 2 of x+5-9

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