Solve for x e^(x+5)=2

$e^{x + 5} = 2$

Step 1

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

$\ln (e^{x + 5}) = \ln (2)$

Step 2

Expand the left side.

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

$(x + 5) \ln (e) = \ln (2)$

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

$(x + 5) \cdot 1 = \ln (2)$

Multiply $x + 5$ by $1$ .

$x + 5 = \ln (2)$

Step 3

Subtract $5$ from both sides of the equation.

$x = \ln (2) - 5$

Step 4

The result can be shown in multiple forms.

Exact Form:

$x = \ln (2) - 5$

Decimal Form:

$x = - 4.30685281 \dots$