Find the Inverse 2^(x-1)+1

$2^{x - 1} + 1$

Step 1

Interchange the variables.

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

Step 2

Solve for $y$ .

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

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

Subtract $1$ from both sides of the equation.

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

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

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

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

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

Simplify the left side.

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

Apply the distributive property.

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

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

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

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

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

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

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

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

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

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

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

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

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

Simplify the left side.

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

Cancel the common factor.

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

Divide $y$ by $1$ .

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

Simplify the right side.

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

Cancel the common factor.

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

Rewrite the expression.

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

Step 3

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

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

Step 4

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

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

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

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

Subtract $1$ from $1$ .

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

Add $2^{x - 1}$ and $0$ .

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

Simplify each term.

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

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

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

Cancel the common factor.

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

Divide $x - 1$ by $1$ .

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

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

Subtract $1$ from $1$ .

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

Add $x$ and $0$ .

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

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

Set up the composite result function.

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

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

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

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

Subtract $1$ from $1$ .

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

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

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

Simplify each term.

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

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

Exponentiation and log are inverse functions.

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

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

Add $- 1$ and $1$ .

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

Add $x$ and $0$ .

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

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

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

Find the Inverse 2^(x-1)+1

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