Find the Inverse 1/2 log of 4x

$\frac{1}{2} \log (4 x)$

Step 1

Interchange the variables.

$x = \frac{1}{2} \cdot \log (4 y)$

Step 2

Solve for $y$ .

Rewrite the equation as $\frac{1}{2} \cdot \log (4 y) = x$ .

$\frac{1}{2} \cdot \log (4 y) = x$

Multiply both sides of the equation by $2$ .

$2 (\frac{1}{2} \cdot \log (4 y)) = 2 x$

Simplify the left side.

Simplify $2 (\frac{1}{2} \cdot \log (4 y))$ .

Combine $\frac{1}{2}$ and $\log (4 y)$ .

$2 \frac{\log (4 y)}{2} = 2 x$

Cancel the common factor of $2$ .

Cancel the common factor.

$2 \frac{\log (4 y)}{2} = 2 x$

Rewrite the expression.

$\log (4 y) = 2 x$

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

$10^{2 x} = 4 y$

Solve for $y$ .

Rewrite the equation as $4 y = 10^{2 x}$ .

$4 y = 10^{2 x}$

Divide each term in $4 y = 10^{2 x}$ by $4$ and simplify.

Divide each term in $4 y = 10^{2 x}$ by $4$ .

$\frac{4 y}{4} = \frac{10^{2 x}}{4}$

Simplify the left side.

Cancel the common factor of $4$ .

Cancel the common factor.

$\frac{4 y}{4} = \frac{10^{2 x}}{4}$

Divide $y$ by $1$ .

$y = \frac{10^{2 x}}{4}$

Step 3

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

$f^{- 1} (x) = \frac{10^{2 x}}{4}$

Step 4

Verify if $f^{- 1} (x) = \frac{10^{2 x}}{4}$ is the inverse of $f (x) = \frac{1}{2} \cdot \log (4 x)$ .

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

$f^{- 1} (\frac{1}{2} \cdot \log (4 x)) = \frac{10^{2 (\frac{1}{2} \cdot \log (4 x))}}{4}$

Simplify the numerator.

Cancel the common factor of $2$ .

Cancel the common factor.

$f^{- 1} (\frac{1}{2} \cdot \log (4 x)) = \frac{10^{2 (\frac{1}{2}) \cdot \log (4 x)}}{4}$

Rewrite the expression.

$f^{- 1} (\frac{1}{2} \cdot \log (4 x)) = \frac{10^{1 \cdot \log (4 x)}}{4}$

Simplify $1 \cdot \log (4 x)$ by moving $1$ inside the logarithm.

$f^{- 1} (\frac{1}{2} \cdot \log (4 x)) = \frac{10^{\log ({(4 x)}^{1})}}{4}$

Exponentiation and log are inverse functions.

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

Simplify.

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

Cancel the common factor of $4$ .

Cancel the common factor.

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

Divide $x$ by $1$ .

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

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

Set up the composite result function.

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

Evaluate $f (\frac{10^{2 x}}{4})$ by substituting in the value of $f^{- 1}$ into $f$ .

$f (\frac{10^{2 x}}{4}) = \frac{1}{2} \cdot \log (4 (\frac{10^{2 x}}{4}))$

Cancel the common factor of $4$ .

Cancel the common factor.

$f (\frac{10^{2 x}}{4}) = \frac{1}{2} \cdot \log (4 (\frac{10^{2 x}}{4}))$

Rewrite the expression.

$f (\frac{10^{2 x}}{4}) = \frac{1}{2} \cdot \log (10^{2 x})$

Use logarithm rules to move $2 x$ out of the exponent.

$f (\frac{10^{2 x}}{4}) = \frac{1}{2} \cdot (2 x \log (10))$

Logarithm base $10$ of $10$ is $1$ .

$f (\frac{10^{2 x}}{4}) = \frac{1}{2} \cdot (2 x \cdot 1)$

Multiply $2$ by $1$ .

$f (\frac{10^{2 x}}{4}) = \frac{1}{2} \cdot (2 x)$

Cancel the common factor of $2$ .

Factor $2$ out of $2 x$ .

$f (\frac{10^{2 x}}{4}) = \frac{1}{2} \cdot (2 (x))$

Cancel the common factor.

$f (\frac{10^{2 x}}{4}) = \frac{1}{2} \cdot (2 x)$

Rewrite the expression.

$f (\frac{10^{2 x}}{4}) = x$

Since $f^{- 1} (f (x)) = x$ and $f (f^{- 1} (x)) = x$ , then $f^{- 1} (x) = \frac{10^{2 x}}{4}$ is the inverse of $f (x) = \frac{1}{2} \cdot \log (4 x)$ .

$f^{- 1} (x) = \frac{10^{2 x}}{4}$

Find the Inverse 1/2 log of 4x

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