Find the Inverse y = log base 2 of 2x

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

Interchange the variables.

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

Solve for $y$ .

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

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

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

$2^{x} = 2 y$

Solve for $y$ .

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

$2 y = 2^{x}$

Divide each term in $2 y = 2^{x}$ by $2$ and simplify.

Divide each term in $2 y = 2^{x}$ by $2$ .

$\frac{2 y}{2} = \frac{2^{x}}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 y}{2} = \frac{2^{x}}{2}$

Divide $y$ by $1$ .

$y = \frac{2^{x}}{2}$

Simplify the right side.

Cancel the common factor of $2^{x}$ and $2$ .

Factor $2$ out of $2^{x}$ .

$y = \frac{2 \cdot 2^{x - 1}}{2}$

Cancel the common factors.

Factor $2$ out of $2$ .

$y = \frac{2 \cdot 2^{x - 1}}{2 (1)}$

Cancel the common factor.

$y = \frac{2 \cdot 2^{x - 1}}{2 \cdot 1}$

Rewrite the expression.

$y = \frac{2^{x - 1}}{1}$

Divide $2^{x - 1}$ by $1$ .

$y = 2^{x - 1}$

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

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

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

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

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

Remove parentheses.

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

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

Set up the composite result function.

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

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

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

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

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

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

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

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

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

Multiply $x - 1$ by $1$ .

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

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

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

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

Subtract $1$ from $1$ .

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

Add $x$ and $0$ .

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

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

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

Find the Inverse y = log base 2 of 2x

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