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

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

Step 1

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

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

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

Multiply by $1$ .

$e^{2 x} (2 x) + e^{2 x} \cdot 1 = 0$

Factor $e^{2 x}$ out of $e^{2 x} (2 x) + e^{2 x} \cdot 1$ .

$e^{2 x} (2 x + 1) = 0$

Step 2

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

$e^{2 x} = 0$

$2 x + 1 = 0$

Step 3

Set $e^{2 x}$ equal to $0$ and solve for $x$ .

Set $e^{2 x}$ equal to $0$ .

$e^{2 x} = 0$

Solve $e^{2 x} = 0$ for $x$ .

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

$\ln (e^{2 x}) = \ln (0)$

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

Undefined

There is no solution for $e^{2 x} = 0$

No solution

Step 4

Set $2 x + 1$ equal to $0$ and solve for $x$ .

Set $2 x + 1$ equal to $0$ .

$2 x + 1 = 0$

Solve $2 x + 1 = 0$ for $x$ .

Subtract $1$ from both sides of the equation.

$2 x = - 1$

Divide each term in $2 x = - 1$ by $2$ and simplify.

Divide each term in $2 x = - 1$ by $2$ .

$\frac{2 x}{2} = \frac{- 1}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 x}{2} = \frac{- 1}{2}$

Divide $x$ by $1$ .

$x = \frac{- 1}{2}$

Simplify the right side.

Move the negative in front of the fraction.

$x = - \frac{1}{2}$

Step 5

The final solution is all the values that make $e^{2 x} (2 x + 1) = 0$ true.

$x = - \frac{1}{2}$

Step 6

The result can be shown in multiple forms.

Exact Form:

$x = - \frac{1}{2}$

Decimal Form:

$x = - 0.5$

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

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