Solve for c -2e^c+14=-6

$- 2 e^{c} + 14 = - 6$

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

Subtract $14$ from both sides of the equation.

$- 2 e^{c} = - 6 - 14$

Subtract $14$ from $- 6$ .

$- 2 e^{c} = - 20$

Divide each term in $- 2 e^{c} = - 20$ by $- 2$ and simplify.

Divide each term in $- 2 e^{c} = - 20$ by $- 2$ .

$\frac{- 2 e^{c}}{- 2} = \frac{- 20}{- 2}$

Simplify the left side.

Cancel the common factor of $- 2$ .

Cancel the common factor.

$\frac{- 2 e^{c}}{- 2} = \frac{- 20}{- 2}$

Divide $e^{c}$ by $1$ .

$e^{c} = \frac{- 20}{- 2}$

Simplify the right side.

Divide $- 20$ by $- 2$ .

$e^{c} = 10$

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

$\ln (e^{c}) = \ln (10)$

Expand the left side.

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

$c \ln (e) = \ln (10)$

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

$c \cdot 1 = \ln (10)$

Multiply $c$ by $1$ .

$c = \ln (10)$

The result can be shown in multiple forms.

Exact Form:

$c = \ln (10)$

Decimal Form:

$c = 2.30258509 \dots$