Solve for x 1+2e^(x+1)=9

$1 + 2 e^{x + 1} = 9$

Step 1

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

Subtract $1$ from both sides of the equation.

$2 e^{x + 1} = 9 - 1$

Subtract $1$ from $9$ .

$2 e^{x + 1} = 8$

Step 2

Divide each term in $2 e^{x + 1} = 8$ by $2$ and simplify.

Divide each term in $2 e^{x + 1} = 8$ by $2$ .

$\frac{2 e^{x + 1}}{2} = \frac{8}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 e^{x + 1}}{2} = \frac{8}{2}$

Divide $e^{x + 1}$ by $1$ .

$e^{x + 1} = \frac{8}{2}$

Simplify the right side.

Divide $8$ by $2$ .

$e^{x + 1} = 4$

Step 3

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

$\ln (e^{x + 1}) = \ln (4)$

Step 4

Expand the left side.

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

$(x + 1) \ln (e) = \ln (4)$

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

$(x + 1) \cdot 1 = \ln (4)$

Multiply $x + 1$ by $1$ .

$x + 1 = \ln (4)$

Step 5

Subtract $1$ from both sides of the equation.

$x = \ln (4) - 1$

Step 6

The result can be shown in multiple forms.

Exact Form:

$x = \ln (4) - 1$

Decimal Form:

$x = 0.38629436 \dots$