Find the Inverse y=(1/2)^x

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

Interchange the variables.

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

Solve for $y$ .

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

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

Take the natural logarithm of both sides of the equation to remove the variable from the exponent.

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

Expand the left side.

Expand $\ln ({(\frac{1}{2})}^{y})$ by moving $y$ outside the logarithm.

$y \ln (\frac{1}{2}) = \ln (x)$

Rewrite $\ln (\frac{1}{2})$ as $\ln (1) - \ln (2)$ .

$y (\ln (1) - \ln (2)) = \ln (x)$

The natural logarithm of $1$ is $0$ .

$y (0 - \ln (2)) = \ln (x)$

Subtract $\ln (2)$ from $0$ .

$y (- \ln (2)) = \ln (x)$

Simplify the left side.

Reorder factors in $y (- \ln (2))$ .

$- y \ln (2) = \ln (x)$

Divide each term in $- y \ln (2) = \ln (x)$ by $- 1 \ln (2)$ and simplify.

Divide each term in $- y \ln (2) = \ln (x)$ by $- 1 \ln (2)$ .

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

Simplify the left side.

Dividing two negative values results in a positive value.

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

Cancel the common factor of $\ln (2)$ .

Cancel the common factor.

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

Divide $y$ by $1$ .

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

Simplify the right side.

Move the negative in front of the fraction.

$y = - \frac{\ln (x)}{\ln (2)}$

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

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

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

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

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

Simplify the numerator.

Apply the product rule to $\frac{1}{2}$ .

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

One to any power is one.

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

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

Set up the composite result function.

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

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

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

Simplify the expression.

Apply the product rule to $\frac{1}{2}$ .

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

One to any power is one.

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

Simplify the denominator.

Use the change of base rule $\frac{\log_{b} (x)}{\log_{b} (a)} = \log_{a} (x)$ .

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

Simplify $- \log_{2} (x)$ by moving $- 1$ inside the logarithm.

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

Exponentiation and log are inverse functions.

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

Rewrite the expression using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

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

Multiply the numerator by the reciprocal of the denominator.

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

Multiply $x$ by $1$ .

$f (- \frac{\ln (x)}{\ln (2)}) = x$

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

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

Find the Inverse y=(1/2)^x

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