Solve for x 2e^(x-9)=40

$2 e^{x - 9} = 40$

Divide each term by $2$ and simplify.

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Divide each term in $2 e^{x - 9} = 40$ by $2$ .

$\frac{2 e^{x - 9}}{2} = \frac{40}{2}$

Cancel the common factor of $2$ .

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Cancel the common factor.

$\frac{2 e^{x - 9}}{2} = \frac{40}{2}$

Divide $e^{x - 9}$ by $1$ .

$e^{x - 9} = \frac{40}{2}$

Divide $40$ by $2$ .

$e^{x - 9} = 20$

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

$\ln (e^{x - 9}) = \ln (20)$

Expand the left side.

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Expand $\ln (e^{x - 9})$ by moving $x - 9$ outside the logarithm.

$(x - 9) \ln (e) = \ln (20)$

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

$(x - 9) \cdot 1 = \ln (20)$

Multiply $x - 9$ by $1$ .

$x - 9 = \ln (20)$

Add $9$ to both sides of the equation.

$x = \ln (20) + 9$

The result can be shown in multiple forms.

Exact Form:

$x = \ln (20) + 9$

Decimal Form:

$x = 11.99573227 \dots$

Solve for x 2e^(x-9)=40

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