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

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Algebra

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

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

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Subtract $1$ from both sides of the equation.

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

Subtract $1$ from $9$ .

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

Divide each term by $2$ and simplify.

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

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

Cancel the common factor of $2$ .

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Cancel the common factor.

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

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

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

Divide $8$ by $2$ .

$e^{x + 1} = 4$

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

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

Expand the left side.

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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 $1$ .

$x + 1 = \ln (4)$

Subtract $1$ from both sides of the equation.

$x = \ln (4) - 1$

The result can be shown in multiple forms.

Exact Form:

$x = \ln (4) - 1$

Decimal Form:

$x = 0.38629436 \dots$

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