Find the Inverse y=e^(-5x)

$y = e^{- 5 x}$

Step 1

Interchange the variables.

$x = e^{- 5 y}$

Step 2

Solve for $y$ .

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

$e^{- 5 y} = x$

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

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

Expand the left side.

Expand $\ln (e^{- 5 y})$ by moving $- 5 y$ outside the logarithm.

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

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

$- 5 y \cdot 1 = \ln (x)$

Multiply $- 5$ by $1$ .

$- 5 y = \ln (x)$

Divide each term in $- 5 y = \ln (x)$ by $- 5$ and simplify.

Divide each term in $- 5 y = \ln (x)$ by $- 5$ .

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

Simplify the left side.

Cancel the common factor of $- 5$ .

Cancel the common factor.

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

Divide $y$ by $1$ .

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

Simplify the right side.

Move the negative in front of the fraction.

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

Step 3

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

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

Step 4

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

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

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

Expand $\ln (e^{- 5 x})$ by moving $- 5 x$ outside the logarithm.

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

Cancel the common factor of $- 5$ and $5$ .

Factor $5$ out of $- 5 x \ln (e)$ .

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

Cancel the common factors.

Factor $5$ out of $5$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

Divide $- x \ln (e)$ by $1$ .

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

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

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

Multiply $- 1$ by $1$ .

$f^{- 1} (e^{- 5 x}) = x$

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

Set up the composite result function.

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

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

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

Cancel the common factor of $5$ .

Move the leading negative in $- \frac{\ln (x)}{5}$ into the numerator.

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

Factor $5$ out of $- 5$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

Multiply.

Multiply $- 1$ by $- 1$ .

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

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

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

Exponentiation and log are inverse functions.

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

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

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

Find the Inverse y=e^(-5x)

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