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

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

Interchange the variables.

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

Solve for $y$ .

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

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

Simplify each term.

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

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

One to any power is one.

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

Subtract $2$ from both sides of the equation.

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

Multiply both sides of the equation by $2^{y - 1}$ .

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

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

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

Divide each term in $2^{y - 1} (x - 2) = 1$ by $x - 2$ and simplify.

Divide each term in $2^{y - 1} (x - 2) = 1$ by $x - 2$ .

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

Simplify the left side.

Cancel the common factor of $x - 2$ .

Cancel the common factor.

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

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

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

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

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

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

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

Simplify the left side.

Simplify $(y - 1) \ln (2)$ .

Apply the distributive property.

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

Rewrite $- 1 \ln (2)$ as $- \ln (2)$ .

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

Move all the terms containing a logarithm to the left side of the equation.

$y \ln (2) - \ln (2) - \ln (\frac{1}{x - 2}) = 0$

Move all terms not containing $y$ to the right side of the equation.

Add $\ln (2)$ to both sides of the equation.

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

Add $\ln (\frac{1}{x - 2})$ to both sides of the equation.

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

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

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

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

Simplify the left side.

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

Cancel the common factor.

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

Divide $y$ by $1$ .

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

Simplify the right side.

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

Cancel the common factor.

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

Rewrite the expression.

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

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

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

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

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

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

Combine the opposite terms in $1 + \frac{\ln (\frac{1}{({(\frac{1}{2})}^{x - 1} + 2) - 2})}{\ln (2)}$ .

Subtract $2$ from $2$ .

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

Add ${(\frac{1}{2})}^{x - 1}$ and $0$ .

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

Simplify each term.

Simplify the denominator.

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

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

One to any power is one.

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

Simplify the numerator.

Multiply the numerator by the reciprocal of the denominator.

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

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

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

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

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

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

Cancel the common factor.

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

Divide $x - 1$ by $1$ .

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

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

Subtract $1$ from $1$ .

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

Add $x$ and $0$ .

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

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

Set up the composite result function.

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

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

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

Combine the opposite terms in ${(\frac{1}{2})}^{(1 + \frac{\ln (\frac{1}{x - 2})}{\ln (2)}) - 1} + 2$ .

Subtract $1$ from $1$ .

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

Add $0$ and $\frac{\ln (\frac{1}{x - 2})}{\ln (2)}$ .

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

Simplify each term.

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

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

One to any power is one.

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

Simplify the denominator.

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

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

Exponentiation and log are inverse functions.

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

Multiply the numerator by the reciprocal of the denominator.

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

Multiply $x - 2$ by $1$ .

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

Combine the opposite terms in $x - 2 + 2$ .

Add $- 2$ and $2$ .

$f (1 + \frac{\ln (\frac{1}{x - 2})}{\ln (2)}) = x + 0$

Add $x$ and $0$ .

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

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

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

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

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