Solve for k 2e^(-k-1)=44

$2 e^{- k - 1} = 44$

Divide each term in $2 e^{- k - 1} = 44$ by $2$ and simplify.

Divide each term in $2 e^{- k - 1} = 44$ by $2$ .

$\frac{2 e^{- k - 1}}{2} = \frac{44}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 e^{- k - 1}}{2} = \frac{44}{2}$

Divide $e^{- k - 1}$ by $1$ .

$e^{- k - 1} = \frac{44}{2}$

Simplify the right side.

Divide $44$ by $2$ .

$e^{- k - 1} = 22$

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

$\ln (e^{- k - 1}) = \ln (22)$

Expand the left side.

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

$(- k - 1) \ln (e) = \ln (22)$

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

$(- k - 1) \cdot 1 = \ln (22)$

Multiply $- k - 1$ by $1$ .

$- k - 1 = \ln (22)$

Add $1$ to both sides of the equation.

$- k = \ln (22) + 1$

Divide each term in $- k = \ln (22) + 1$ by $- 1$ and simplify.

Divide each term in $- k = \ln (22) + 1$ by $- 1$ .

$\frac{- k}{- 1} = \frac{\ln (22)}{- 1} + \frac{1}{- 1}$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{k}{1} = \frac{\ln (22)}{- 1} + \frac{1}{- 1}$

Divide $k$ by $1$ .

$k = \frac{\ln (22)}{- 1} + \frac{1}{- 1}$

Simplify the right side.

Simplify each term.

Move the negative one from the denominator of $\frac{\ln (22)}{- 1}$ .

$k = - 1 \cdot \ln (22) + \frac{1}{- 1}$

Rewrite $- 1 \cdot \ln (22)$ as $- \ln (22)$ .

$k = - \ln (22) + \frac{1}{- 1}$

Divide $1$ by $- 1$ .

$k = - \ln (22) - 1$

The result can be shown in multiple forms.

Exact Form:

$k = - \ln (22) - 1$

Decimal Form:

$k = - 4.09104245 \dots$