Find the Inverse y=(1/4)^x

$y = {(\frac{1}{4})}^{x}$

Step 1

Interchange the variables.

$x = {(\frac{1}{4})}^{y}$

Step 2

Solve for $y$ .

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

${(\frac{1}{4})}^{y} = x$

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

$\ln ({(\frac{1}{4})}^{y}) = \ln (x)$

Expand the left side.

Expand $\ln ({(\frac{1}{4})}^{y})$ by moving $y$ outside the logarithm.

$y \ln (\frac{1}{4}) = \ln (x)$

Rewrite $\ln (\frac{1}{4})$ as $\ln (1) - \ln (4)$ .

$y (\ln (1) - \ln (4)) = \ln (x)$

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

$y (0 - \ln (4)) = \ln (x)$

Rewrite $\ln (4)$ as $\ln (2^{2})$ .

$y (0 - \ln (2^{2})) = \ln (x)$

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

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

Multiply $2$ by $- 1$ .

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

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

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

Simplify the left side.

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

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

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

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

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

Simplify the left side.

Cancel the common factor of $- 2$ .

Cancel the common factor.

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

Rewrite the expression.

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

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

Cancel the common factor.

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

Divide $y$ by $1$ .

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

Simplify the right side.

Move the negative in front of the fraction.

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

Step 3

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

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

Step 4

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

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

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

Simplify the numerator.

Apply the product rule to $\frac{1}{4}$ .

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

One to any power is one.

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

Simplify $2 \ln (2)$ by moving $2$ inside the logarithm.

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

Raise $2$ to the power of $2$ .

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

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

Set up the composite result function.

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

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

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

Simplify $2 \ln (2)$ by moving $2$ inside the logarithm.

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

Simplify the expression.

Raise $2$ to the power of $2$ .

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

Apply the product rule to $\frac{1}{4}$ .

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

One to any power is one.

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

Simplify the denominator.

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

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

Simplify $- \log_{4} (x)$ by moving $- 1$ inside the logarithm.

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

Exponentiation and log are inverse functions.

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

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

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

Multiply the numerator by the reciprocal of the denominator.

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

Multiply $x$ by $1$ .

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

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

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

Find the Inverse y=(1/4)^x

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