Find the Inverse log base 2 of x-1-1

$\log_{2} (x - 1) - 1$

Interchange the variables.

$x = \log_{2} (y - 1) - 1$

Solve for $y$ .

Rewrite the equation as $\log_{2} (y - 1) - 1 = x$ .

$\log_{2} (y - 1) - 1 = x$

Add $1$ to both sides of the equation.

$\log_{2} (y - 1) = x + 1$

Rewrite $\log_{2} (y - 1) = x + 1$ 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 $b^{y} = x$ .

$2^{x + 1} = y - 1$

Solve for $y$ .

Rewrite the equation as $y - 1 = 2^{x + 1}$ .

$y - 1 = 2^{x + 1}$

Add $1$ to both sides of the equation.

$y = 2^{x + 1} + 1$

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

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

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

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

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

Combine the opposite terms in $2^{(\log_{2} (x - 1) - 1) + 1} + 1$ .

Add $- 1$ and $1$ .

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

Add $\log_{2} (x - 1)$ and $0$ .

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

Exponentiation and log are inverse functions.

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

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

Add $- 1$ and $1$ .

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

Add $x$ and $0$ .

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

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

Set up the composite result function.

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

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

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

Combine the opposite terms in $\log_{2} ((2^{x + 1} + 1) - 1) - 1$ .

Subtract $1$ from $1$ .

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

Add $2^{x + 1}$ and $0$ .

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

Simplify each term.

Use logarithm rules to move $x + 1$ out of the exponent.

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

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

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

Multiply $x + 1$ by $1$ .

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

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

Subtract $1$ from $1$ .

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

Add $x$ and $0$ .

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

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

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

Find the Inverse log base 2 of x-1-1

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