Find the Inverse f(x)=4+5/(e^(x-3))

$f (x) = 4 + \frac{5}{e^{x - 3}}$

Write $f (x) = 4 + \frac{5}{e^{x - 3}}$ as an equation.

$y = 4 + \frac{5}{e^{x - 3}}$

Interchange the variables.

$x = 4 + \frac{5}{e^{y - 3}}$

Solve for $y$ .

Rewrite the equation as $4 + \frac{5}{e^{y - 3}} = x$ .

$4 + \frac{5}{e^{y - 3}} = x$

Subtract $4$ from both sides of the equation.

$\frac{5}{e^{y - 3}} = x - 4$

Multiply both sides by $e^{y - 3}$ .

$\frac{5}{e^{y - 3}} e^{y - 3} = (x - 4) e^{y - 3}$

Simplify.

Simplify the left side.

Cancel the common factor of $e^{y - 3}$ .

Cancel the common factor.

$\frac{5}{e^{y - 3}} e^{y - 3} = (x - 4) e^{y - 3}$

Rewrite the expression.

$5 = (x - 4) e^{y - 3}$

Simplify the right side.

Apply the distributive property.

$5 = x e^{y - 3} - 4 e^{y - 3}$

Solve for $y$ .

Rewrite the equation as $x e^{y - 3} - 4 e^{y - 3} = 5$ .

$x e^{y - 3} - 4 e^{y - 3} = 5$

Factor $e^{y - 3}$ out of $x e^{y - 3} - 4 e^{y - 3}$ .

Factor $e^{y - 3}$ out of $x e^{y - 3}$ .

$e^{y - 3} x - 4 e^{y - 3} = 5$

Factor $e^{y - 3}$ out of $- 4 e^{y - 3}$ .

$e^{y - 3} x + e^{y - 3} \cdot - 4 = 5$

Factor $e^{y - 3}$ out of $e^{y - 3} x + e^{y - 3} \cdot - 4$ .

$e^{y - 3} (x - 4) = 5$

Divide each term in $e^{y - 3} (x - 4) = 5$ by $x - 4$ and simplify.

Divide each term in $e^{y - 3} (x - 4) = 5$ by $x - 4$ .

$\frac{e^{y - 3} (x - 4)}{x - 4} = \frac{5}{x - 4}$

Simplify the left side.

Cancel the common factor of $x - 4$ .

Cancel the common factor.

$\frac{e^{y - 3} (x - 4)}{x - 4} = \frac{5}{x - 4}$

Divide $e^{y - 3}$ by $1$ .

$e^{y - 3} = \frac{5}{x - 4}$

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

$\ln (e^{y - 3}) = \ln (\frac{5}{x - 4})$

Expand the left side.

Expand $\ln (e^{y - 3})$ by moving $y - 3$ outside the logarithm.

$(y - 3) \ln (e) = \ln (\frac{5}{x - 4})$

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

$(y - 3) \cdot 1 = \ln (\frac{5}{x - 4})$

Multiply $y - 3$ by $1$ .

$y - 3 = \ln (\frac{5}{x - 4})$

Add $3$ to both sides of the equation.

$y = \ln (\frac{5}{x - 4}) + 3$

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

$f^{- 1} (x) = \ln (\frac{5}{x - 4}) + 3$

Verify if $f^{- 1} (x) = \ln (\frac{5}{x - 4}) + 3$ is the inverse of $f (x) = 4 + \frac{5}{e^{x - 3}}$ .

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 $f^{- 1} (4 + \frac{5}{e^{x - 3}})$ by substituting in the value of $f$ into $f^{- 1}$ .

$f^{- 1} (4 + \frac{5}{e^{x - 3}}) = \ln (\frac{5}{(4 + \frac{5}{e^{x - 3}}) - 4}) + 3$

Combine the opposite terms in $\ln (\frac{5}{(4 + \frac{5}{e^{x - 3}}) - 4}) + 3$ .

Subtract $4$ from $4$ .

$f^{- 1} (4 + \frac{5}{e^{x - 3}}) = \ln (\frac{5}{0 + \frac{5}{e^{x - 3}}}) + 3$

Add $0$ and $\frac{5}{e^{x - 3}}$ .

$f^{- 1} (4 + \frac{5}{e^{x - 3}}) = \ln (\frac{5}{\frac{5}{e^{x - 3}}}) + 3$

Simplify each term.

Multiply the numerator by the reciprocal of the denominator.

$f^{- 1} (4 + \frac{5}{e^{x - 3}}) = \ln (5 (\frac{e^{x - 3}}{5})) + 3$

Cancel the common factor of $5$ .

Cancel the common factor.

$f^{- 1} (4 + \frac{5}{e^{x - 3}}) = \ln (5 (\frac{e^{x - 3}}{5})) + 3$

Rewrite the expression.

$f^{- 1} (4 + \frac{5}{e^{x - 3}}) = \ln (e^{x - 3}) + 3$

Use logarithm rules to move $x - 3$ out of the exponent.

$f^{- 1} (4 + \frac{5}{e^{x - 3}}) = (x - 3) \ln (e) + 3$

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

$f^{- 1} (4 + \frac{5}{e^{x - 3}}) = (x - 3) \cdot 1 + 3$

Multiply $x - 3$ by $1$ .

$f^{- 1} (4 + \frac{5}{e^{x - 3}}) = x - 3 + 3$

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

Add $- 3$ and $3$ .

$f^{- 1} (4 + \frac{5}{e^{x - 3}}) = x + 0$

Add $x$ and $0$ .

$f^{- 1} (4 + \frac{5}{e^{x - 3}}) = x$

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

Set up the composite result function.

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

Evaluate $f (\ln (\frac{5}{x - 4}) + 3)$ by substituting in the value of $f^{- 1}$ into $f$ .

$f (\ln (\frac{5}{x - 4}) + 3) = 4 + \frac{5}{e^{(\ln (\frac{5}{x - 4}) + 3) - 3}}$

Combine the opposite terms in $4 + \frac{5}{e^{(\ln (\frac{5}{x - 4}) + 3) - 3}}$ .

Subtract $3$ from $3$ .

$f (\ln (\frac{5}{x - 4}) + 3) = 4 + \frac{5}{e^{\ln (\frac{5}{x - 4}) + 0}}$

Add $\ln (\frac{5}{x - 4})$ and $0$ .

$f (\ln (\frac{5}{x - 4}) + 3) = 4 + \frac{5}{e^{\ln (\frac{5}{x - 4})}}$

Simplify each term.

Exponentiation and log are inverse functions.

$f (\ln (\frac{5}{x - 4}) + 3) = 4 + \frac{5}{\frac{5}{x - 4}}$

Multiply the numerator by the reciprocal of the denominator.

$f (\ln (\frac{5}{x - 4}) + 3) = 4 + 5 (\frac{x - 4}{5})$

Cancel the common factor of $5$ .

Cancel the common factor.

$f (\ln (\frac{5}{x - 4}) + 3) = 4 + 5 (\frac{x - 4}{5})$

Rewrite the expression.

$f (\ln (\frac{5}{x - 4}) + 3) = 4 + x - 4$

Combine the opposite terms in $4 + x - 4$ .

Subtract $4$ from $4$ .

$f (\ln (\frac{5}{x - 4}) + 3) = x + 0$

Add $x$ and $0$ .

$f (\ln (\frac{5}{x - 4}) + 3) = x$

Since $f^{- 1} (f (x)) = x$ and $f (f^{- 1} (x)) = x$ , then $f^{- 1} (x) = \ln (\frac{5}{x - 4}) + 3$ is the inverse of $f (x) = 4 + \frac{5}{e^{x - 3}}$ .

$f^{- 1} (x) = \ln (\frac{5}{x - 4}) + 3$

Find the Inverse f(x)=4+5/(e^(x-3))

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