Solve for x e^(x+1)=30

$e^{x + 1} = 30$

Step 1

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

$\ln (e^{x + 1}) = \ln (30)$

Step 2

Expand the left side.

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

$(x + 1) \ln (e) = \ln (30)$

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

$(x + 1) \cdot 1 = \ln (30)$

Multiply $x + 1$ by $1$ .

$x + 1 = \ln (30)$

Step 3

Subtract $1$ from both sides of the equation.

$x = \ln (30) - 1$

Step 4

The result can be shown in multiple forms.

Exact Form:

$x = \ln (30) - 1$

Decimal Form:

$x = 2.40119738 \dots$