Solve for x e^(x+1)+1=32

$e^{x + 1} + 1 = 32$

Step 1

Move all terms not containing $x$ to the right side of the equation.

Subtract $1$ from both sides of the equation.

$e^{x + 1} = 32 - 1$

Subtract $1$ from $32$ .

$e^{x + 1} = 31$

Step 2

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

$\ln (e^{x + 1}) = \ln (31)$

Step 3

Expand the left side.

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

$(x + 1) \ln (e) = \ln (31)$

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

$(x + 1) \cdot 1 = \ln (31)$

Multiply $x + 1$ by $1$ .

$x + 1 = \ln (31)$

Step 4

Subtract $1$ from both sides of the equation.

$x = \ln (31) - 1$

Step 5

The result can be shown in multiple forms.

Exact Form:

$x = \ln (31) - 1$

Decimal Form:

$x = 2.43398720 \dots$