Solve for x xe^x+2e^x=0

$x e^{x} + 2 e^{x} = 0$

Step 1

Factor $e^{x}$ out of $x e^{x} + 2 e^{x}$ .

Factor $e^{x}$ out of $x e^{x}$ .

$e^{x} x + 2 e^{x} = 0$

Factor $e^{x}$ out of $2 e^{x}$ .

$e^{x} x + e^{x} \cdot 2 = 0$

Factor $e^{x}$ out of $e^{x} x + e^{x} \cdot 2$ .

$e^{x} (x + 2) = 0$

Step 2

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$e^{x} = 0$

$x + 2 = 0$

Step 3

Set $e^{x}$ equal to $0$ and solve for $x$ .

Set $e^{x}$ equal to $0$ .

$e^{x} = 0$

Solve $e^{x} = 0$ for $x$ .

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

$\ln (e^{x}) = \ln (0)$

The equation cannot be solved because $\ln (0)$ is undefined.

Undefined

There is no solution for $e^{x} = 0$

No solution

Step 4

Set $x + 2$ equal to $0$ and solve for $x$ .

Set $x + 2$ equal to $0$ .

$x + 2 = 0$

Subtract $2$ from both sides of the equation.

$x = - 2$

Step 5

The final solution is all the values that make $e^{x} (x + 2) = 0$ true.

$x = - 2$