Solve for x e^(-x)=5

$e^{- x} = 5$

Step 1

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

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

Step 2

Expand the left side.

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

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

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

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

Multiply $- 1$ by $1$ .

$- x = \ln (5)$

Step 3

Divide each term in $- x = \ln (5)$ by $- 1$ and simplify.

Divide each term in $- x = \ln (5)$ by $- 1$ .

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

Simplify the left side.

Dividing two negative values results in a positive value.

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

Divide $x$ by $1$ .

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

Simplify the right side.

Move the negative one from the denominator of $\frac{\ln (5)}{- 1}$ .

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

Rewrite $- 1 \cdot \ln (5)$ as $- \ln (5)$ .

$x = - \ln (5)$

Step 4

The result can be shown in multiple forms.

Exact Form:

$x = - \ln (5)$

Decimal Form:

$x = - 1.60943791 \dots$