Find the Inverse y=1/5*e^(x+2)

$y = \frac{1}{5} \cdot e^{x + 2}$

Step 1

Interchange the variables.

$x = \frac{1}{5} \cdot e^{y + 2}$

Step 2

Solve for $y$ .

Rewrite the equation as $\frac{1}{5} \cdot e^{y + 2} = x$ .

$\frac{1}{5} \cdot e^{y + 2} = x$

Multiply both sides of the equation by $5$ .

$5 (\frac{1}{5} \cdot e^{y + 2}) = 5 x$

Simplify the left side.

Simplify $5 (\frac{1}{5} \cdot e^{y + 2})$ .

Combine $\frac{1}{5}$ and $e^{y + 2}$ .

$5 \frac{e^{y + 2}}{5} = 5 x$

Cancel the common factor of $5$ .

Cancel the common factor.

$5 \frac{e^{y + 2}}{5} = 5 x$

Rewrite the expression.

$e^{y + 2} = 5 x$

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

$\ln (e^{y + 2}) = \ln (5 x)$

Expand the left side.

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

$(y + 2) \ln (e) = \ln (5 x)$

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

$(y + 2) \cdot 1 = \ln (5 x)$

Multiply $y + 2$ by $1$ .

$y + 2 = \ln (5 x)$

Subtract $2$ from both sides of the equation.

$y = \ln (5 x) - 2$

Step 3

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

$f^{- 1} (x) = \ln (5 x) - 2$

Step 4

Verify if $f^{- 1} (x) = \ln (5 x) - 2$ is the inverse of $f (x) = \frac{1}{5} \cdot e^{x + 2}$ .

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 $f^{- 1} (\frac{1}{5} \cdot e^{x + 2})$ by substituting in the value of $f$ into $f^{- 1}$ .

$f^{- 1} (\frac{1}{5} \cdot e^{x + 2}) = \ln (5 (\frac{1}{5} \cdot e^{x + 2})) - 2$

Simplify each term.

Combine $\frac{1}{5}$ and $e^{x + 2}$ .

$f^{- 1} (\frac{1}{5} \cdot e^{x + 2}) = \ln (5 (\frac{e^{x + 2}}{5})) - 2$

Cancel the common factor of $5$ .

Cancel the common factor.

$f^{- 1} (\frac{1}{5} \cdot e^{x + 2}) = \ln (5 (\frac{e^{x + 2}}{5})) - 2$

Rewrite the expression.

$f^{- 1} (\frac{1}{5} \cdot e^{x + 2}) = \ln (e^{x + 2}) - 2$

Use logarithm rules to move $x + 2$ out of the exponent.

$f^{- 1} (\frac{1}{5} \cdot e^{x + 2}) = (x + 2) \ln (e) - 2$

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

$f^{- 1} (\frac{1}{5} \cdot e^{x + 2}) = (x + 2) \cdot 1 - 2$

Multiply $x + 2$ by $1$ .

$f^{- 1} (\frac{1}{5} \cdot e^{x + 2}) = x + 2 - 2$

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

Subtract $2$ from $2$ .

$f^{- 1} (\frac{1}{5} \cdot e^{x + 2}) = x + 0$

Add $x$ and $0$ .

$f^{- 1} (\frac{1}{5} \cdot e^{x + 2}) = x$

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

Set up the composite result function.

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

Evaluate $f (\ln (5 x) - 2)$ by substituting in the value of $f^{- 1}$ into $f$ .

$f (\ln (5 x) - 2) = \frac{1}{5} \cdot e^{(\ln (5 x) - 2) + 2}$

Combine the opposite terms in $(\ln (5 x) - 2) + 2$ .

Add $- 2$ and $2$ .

$f (\ln (5 x) - 2) = \frac{1}{5} \cdot e^{\ln (5 x) + 0}$

Add $\ln (5 x)$ and $0$ .

$f (\ln (5 x) - 2) = \frac{1}{5} \cdot e^{\ln (5 x)}$

Exponentiation and log are inverse functions.

$f (\ln (5 x) - 2) = \frac{1}{5} \cdot (5 x)$

Cancel the common factor of $5$ .

Factor $5$ out of $5 x$ .

$f (\ln (5 x) - 2) = \frac{1}{5} \cdot (5 (x))$

Cancel the common factor.

$f (\ln (5 x) - 2) = \frac{1}{5} \cdot (5 x)$

Rewrite the expression.

$f (\ln (5 x) - 2) = x$

Since $f^{- 1} (f (x)) = x$ and $f (f^{- 1} (x)) = x$ , then $f^{- 1} (x) = \ln (5 x) - 2$ is the inverse of $f (x) = \frac{1}{5} \cdot e^{x + 2}$ .

$f^{- 1} (x) = \ln (5 x) - 2$

Find the Inverse y=1/5*e^(x+2)

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