Find the Inverse y=e^(8-x)

$y = e^{8 - x}$

Interchange the variables.

$x = e^{8 - y}$

Solve for $y$ .

Rewrite the equation as $e^{8 - y} = x$ .

$e^{8 - y} = x$

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

$\ln (e^{8 - y}) = \ln (x)$

Expand the left side.

Expand $\ln (e^{8 - y})$ by moving $8 - y$ outside the logarithm.

$(8 - y) \ln (e) = \ln (x)$

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

$(8 - y) \cdot 1 = \ln (x)$

Multiply $8 - y$ by $1$ .

$8 - y = \ln (x)$

Subtract $8$ from both sides of the equation.

$- y = \ln (x) - 8$

Divide each term in $- y = \ln (x) - 8$ by $- 1$ and simplify.

Divide each term in $- y = \ln (x) - 8$ by $- 1$ .

$\frac{- y}{- 1} = \frac{\ln (x)}{- 1} + \frac{- 8}{- 1}$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{y}{1} = \frac{\ln (x)}{- 1} + \frac{- 8}{- 1}$

Divide $y$ by $1$ .

$y = \frac{\ln (x)}{- 1} + \frac{- 8}{- 1}$

Simplify the right side.

Simplify each term.

Move the negative one from the denominator of $\frac{\ln (x)}{- 1}$ .

$y = - 1 \cdot \ln (x) + \frac{- 8}{- 1}$

Rewrite $- 1 \cdot \ln (x)$ as $- \ln (x)$ .

$y = - \ln (x) + \frac{- 8}{- 1}$

Divide $- 8$ by $- 1$ .

$y = - \ln (x) + 8$

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

$f^{- 1} (x) = - \ln (x) + 8$

Verify if $f^{- 1} (x) = - \ln (x) + 8$ is the inverse of $f (x) = e^{8 - x}$ .

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 $f^{- 1} (e^{8 - x})$ by substituting in the value of $f$ into $f^{- 1}$ .

$f^{- 1} (e^{8 - x}) = - \ln (e^{8 - x}) + 8$

Simplify each term.

Use logarithm rules to move $8 - x$ out of the exponent.

$f^{- 1} (e^{8 - x}) = - ((8 - x) \ln (e)) + 8$

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

$f^{- 1} (e^{8 - x}) = - ((8 - x) \cdot 1) + 8$

Multiply $8 - x$ by $1$ .

$f^{- 1} (e^{8 - x}) = - (8 - x) + 8$

Apply the distributive property.

$f^{- 1} (e^{8 - x}) = - 1 \cdot 8 + x + 8$

Multiply $- 1$ by $8$ .

$f^{- 1} (e^{8 - x}) = - 8 + x + 8$

Multiply $- - x$ .

Multiply $- 1$ by $- 1$ .

$f^{- 1} (e^{8 - x}) = - 8 + 1 x + 8$

Multiply $x$ by $1$ .

$f^{- 1} (e^{8 - x}) = - 8 + x + 8$

Combine the opposite terms in $- 8 + x + 8$ .

Add $- 8$ and $8$ .

$f^{- 1} (e^{8 - x}) = x + 0$

Add $x$ and $0$ .

$f^{- 1} (e^{8 - x}) = x$

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

Set up the composite result function.

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

Evaluate $f (- \ln (x) + 8)$ by substituting in the value of $f^{- 1}$ into $f$ .

$f (- \ln (x) + 8) = e^{8 - (- \ln (x) + 8)}$

Simplify each term.

Apply the distributive property.

$f (- \ln (x) + 8) = e^{8 + \ln (x) - 1 \cdot 8}$

Multiply $- - \ln (x)$ .

Multiply $- 1$ by $- 1$ .

$f (- \ln (x) + 8) = e^{8 + 1 \ln (x) - 1 \cdot 8}$

Multiply $\ln (x)$ by $1$ .

$f (- \ln (x) + 8) = e^{8 + \ln (x) - 1 \cdot 8}$

Multiply $- 1$ by $8$ .

$f (- \ln (x) + 8) = e^{8 + \ln (x) - 8}$

Combine the opposite terms in $8 + \ln (x) - 8$ .

Subtract $8$ from $8$ .

$f (- \ln (x) + 8) = e^{0 + \ln (x)}$

Add $0$ and $\ln (x)$ .

$f (- \ln (x) + 8) = e^{\ln (x)}$

Exponentiation and log are inverse functions.

$f (- \ln (x) + 8) = x$

Since $f^{- 1} (f (x)) = x$ and $f (f^{- 1} (x)) = x$ , then $f^{- 1} (x) = - \ln (x) + 8$ is the inverse of $f (x) = e^{8 - x}$ .

$f^{- 1} (x) = - \ln (x) + 8$

Find the Inverse y=e^(8-x)

Related Questions

Support LCMGCF.com by disabling your adblocker.