Solve for x 4e^(-2x)=17

$4 e^{- 2 x} = 17$

Step 1

Divide each term in $4 e^{- 2 x} = 17$ by $4$ and simplify.

Divide each term in $4 e^{- 2 x} = 17$ by $4$ .

$\frac{4 e^{- 2 x}}{4} = \frac{17}{4}$

Simplify the left side.

Cancel the common factor of $4$ .

Cancel the common factor.

$\frac{4 e^{- 2 x}}{4} = \frac{17}{4}$

Divide $e^{- 2 x}$ by $1$ .

$e^{- 2 x} = \frac{17}{4}$

Step 2

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

$\ln (e^{- 2 x}) = \ln (\frac{17}{4})$

Step 3

Expand the left side.

Expand $\ln (e^{- 2 x})$ by moving $- 2 x$ outside the logarithm.

$- 2 x \ln (e) = \ln (\frac{17}{4})$

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

$- 2 x \cdot 1 = \ln (\frac{17}{4})$

Multiply $- 2$ by $1$ .

$- 2 x = \ln (\frac{17}{4})$

Step 4

Divide each term in $- 2 x = \ln (\frac{17}{4})$ by $- 2$ and simplify.

Divide each term in $- 2 x = \ln (\frac{17}{4})$ by $- 2$ .

$\frac{- 2 x}{- 2} = \frac{\ln (\frac{17}{4})}{- 2}$

Simplify the left side.

Cancel the common factor of $- 2$ .

Cancel the common factor.

$\frac{- 2 x}{- 2} = \frac{\ln (\frac{17}{4})}{- 2}$

Divide $x$ by $1$ .

$x = \frac{\ln (\frac{17}{4})}{- 2}$

Simplify the right side.

Move the negative in front of the fraction.

$x = - \frac{\ln (\frac{17}{4})}{2}$

Step 5

The result can be shown in multiple forms.

Exact Form:

$x = - \frac{\ln (\frac{17}{4})}{2}$

Decimal Form:

$x = - 0.72345949 \dots$