Solve for x e^xe^(x+1)=1

$e^{x} e^{x + 1} = 1$

Step 1

Multiply $e^{x}$ by $e^{x + 1}$ by adding the exponents.

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$e^{x + x + 1} = 1$

Add $x$ and $x$ .

$e^{2 x + 1} = 1$

Step 2

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

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

Step 3

Expand the left side.

Expand $\ln (e^{2 x + 1})$ by moving $2 x + 1$ outside the logarithm.

$(2 x + 1) \ln (e) = \ln (1)$

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

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

Multiply $2 x + 1$ by $1$ .

$2 x + 1 = \ln (1)$

Step 4

Simplify the right side.

The natural logarithm of $1$ is $0$ .

$2 x + 1 = 0$

Step 5

Subtract $1$ from both sides of the equation.

$2 x = - 1$

Step 6

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 7

The result can be shown in multiple forms.

Exact Form:

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

Decimal Form:

$x = - 0.5$