Find the Inverse y+3=2^x

$y + 3 = 2^{x}$

Subtract $3$ from both sides of the equation.

$y = 2^{x} - 3$

Interchange the variables.

$x = 2^{y} - 3$

Solve for $y$ .

Rewrite the equation as $2^{y} - 3 = x$ .

$2^{y} - 3 = x$

Add $3$ to both sides of the equation.

$2^{y} = x + 3$

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

$\ln (2^{y}) = \ln (x + 3)$

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

$y \ln (2) = \ln (x + 3)$

Divide each term in $y \ln (2) = \ln (x + 3)$ by $\ln (2)$ and simplify.

Divide each term in $y \ln (2) = \ln (x + 3)$ by $\ln (2)$ .

$\frac{y \ln (2)}{\ln (2)} = \frac{\ln (x + 3)}{\ln (2)}$

Simplify the left side.

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

Cancel the common factor.

$\frac{y \ln (2)}{\ln (2)} = \frac{\ln (x + 3)}{\ln (2)}$

Divide $y$ by $1$ .

$y = \frac{\ln (x + 3)}{\ln (2)}$

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

$f^{- 1} (x) = \frac{\ln (x + 3)}{\ln (2)}$

Verify if $f^{- 1} (x) = \frac{\ln (x + 3)}{\ln (2)}$ is the inverse of $f (x) = 2^{x} - 3$ .

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 $f^{- 1} (2^{x} - 3)$ by substituting in the value of $f$ into $f^{- 1}$ .

$f^{- 1} (2^{x} - 3) = \frac{\ln ((2^{x} - 3) + 3)}{\ln (2)}$

Combine the opposite terms in $2^{x} - 3 + 3$ .

Add $- 3$ and $3$ .

$f^{- 1} (2^{x} - 3) = \frac{\ln (2^{x} + 0)}{\ln (2)}$

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

$f^{- 1} (2^{x} - 3) = \frac{\ln (2^{x})}{\ln (2)}$

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

$f^{- 1} (2^{x} - 3) = \frac{x \ln (2)}{\ln (2)}$

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

Cancel the common factor.

$f^{- 1} (2^{x} - 3) = \frac{x \ln (2)}{\ln (2)}$

Divide $x$ by $1$ .

$f^{- 1} (2^{x} - 3) = x$

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

Set up the composite result function.

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

Evaluate $f (\frac{\ln (x + 3)}{\ln (2)})$ by substituting in the value of $f^{- 1}$ into $f$ .

$f (\frac{\ln (x + 3)}{\ln (2)}) = 2^{\frac{\ln (x + 3)}{\ln (2)}} - 3$

Simplify each term.

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

$f (\frac{\ln (x + 3)}{\ln (2)}) = 2^{\log_{2} (x + 3)} - 3$

Exponentiation and log are inverse functions.

$f (\frac{\ln (x + 3)}{\ln (2)}) = x + 3 - 3$

Combine the opposite terms in $x + 3 - 3$ .

Subtract $3$ from $3$ .

$f (\frac{\ln (x + 3)}{\ln (2)}) = x + 0$

Add $x$ and $0$ .

$f (\frac{\ln (x + 3)}{\ln (2)}) = x$

Since $f^{- 1} (f (x)) = x$ and $f (f^{- 1} (x)) = x$ , then $f^{- 1} (x) = \frac{\ln (x + 3)}{\ln (2)}$ is the inverse of $f (x) = 2^{x} - 3$ .

$f^{- 1} (x) = \frac{\ln (x + 3)}{\ln (2)}$

Find the Inverse y+3=2^x

Related Questions

Support LCMGCF.com by disabling your adblocker.