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 $y$ .

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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 $y$

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

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

Add $1$ to both sides of the equation.

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

Solve for $y$ and replace with $f^{- 1} (x)$ .

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

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

Set up the composite result function.

$g (f (x))$

Evaluate $g (f (x))$ by substituting in the value of $f$ into $g$ .

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

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

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Add $- 1$ and $1$ .

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

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

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

Exponentiation and log are inverse functions.

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

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

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Add $- 1$ and $1$ .

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

Add $x$ and $0$ .

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

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