Find the Inverse y = log base 2 of 4x

$y = \log_{2} (4 x)$

Step 1

Interchange the variables.

$x = \log_{2} (4 y)$

Step 2

Solve for $y$ .

Rewrite the equation as $\log_{2} (4 y) = x$ .

$\log_{2} (4 y) = x$

Rewrite $\log_{2} (4 y) = x$ 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 $b^{y} = x$ .

$2^{x} = 4 y$

Solve for $y$ .

Rewrite the equation as $4 y = 2^{x}$ .

$4 y = 2^{x}$

Divide each term in $4 y = 2^{x}$ by $4$ and simplify.

Divide each term in $4 y = 2^{x}$ by $4$ .

$\frac{4 y}{4} = \frac{2^{x}}{4}$

Simplify the left side.

Cancel the common factor of $4$ .

Cancel the common factor.

$\frac{4 y}{4} = \frac{2^{x}}{4}$

Divide $y$ by $1$ .

$y = \frac{2^{x}}{4}$

Step 3

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

$f^{- 1} (x) = \frac{2^{x}}{4}$

Step 4

Verify if $f^{- 1} (x) = \frac{2^{x}}{4}$ is the inverse of $f (x) = \log_{2} (4 x)$ .

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 $f^{- 1} (\log_{2} (4 x))$ by substituting in the value of $f$ into $f^{- 1}$ .

$f^{- 1} (\log_{2} (4 x)) = \frac{2^{\log_{2} (4 x)}}{4}$

Exponentiation and log are inverse functions.

$f^{- 1} (\log_{2} (4 x)) = \frac{4 x}{4}$

Cancel the common factor of $4$ .

Cancel the common factor.

$f^{- 1} (\log_{2} (4 x)) = \frac{4 x}{4}$

Divide $x$ by $1$ .

$f^{- 1} (\log_{2} (4 x)) = x$

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

Set up the composite result function.

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

Evaluate $f (\frac{2^{x}}{4})$ by substituting in the value of $f^{- 1}$ into $f$ .

$f (\frac{2^{x}}{4}) = \log_{2} (4 (\frac{2^{x}}{4}))$

Cancel the common factor of $4$ .

Cancel the common factor.

$f (\frac{2^{x}}{4}) = \log_{2} (4 (\frac{2^{x}}{4}))$

Rewrite the expression.

$f (\frac{2^{x}}{4}) = \log_{2} (2^{x})$

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

$f (\frac{2^{x}}{4}) = x \log_{2} (2)$

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

$f (\frac{2^{x}}{4}) = x \cdot 1$

Multiply $x$ by $1$ .

$f (\frac{2^{x}}{4}) = x$

Since $f^{- 1} (f (x)) = x$ and $f (f^{- 1} (x)) = x$ , then $f^{- 1} (x) = \frac{2^{x}}{4}$ is the inverse of $f (x) = \log_{2} (4 x)$ .

$f^{- 1} (x) = \frac{2^{x}}{4}$

Find the Inverse y = log base 2 of 4x

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