Solve for a e^(-10a-4)=45

$e^{- 10 a - 4} = 45$

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

$\ln (e^{- 10 a - 4}) = \ln (45)$

Expand the left side.

Expand $\ln (e^{- 10 a - 4})$ by moving $- 10 a - 4$ outside the logarithm.

$(- 10 a - 4) \ln (e) = \ln (45)$

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

$(- 10 a - 4) \cdot 1 = \ln (45)$

Multiply $- 10 a - 4$ by $1$ .

$- 10 a - 4 = \ln (45)$

Add $4$ to both sides of the equation.

$- 10 a = \ln (45) + 4$

Divide each term in $- 10 a = \ln (45) + 4$ by $- 10$ and simplify.

Divide each term in $- 10 a = \ln (45) + 4$ by $- 10$ .

$\frac{- 10 a}{- 10} = \frac{\ln (45)}{- 10} + \frac{4}{- 10}$

Simplify the left side.

Cancel the common factor of $- 10$ .

Cancel the common factor.

$\frac{- 10 a}{- 10} = \frac{\ln (45)}{- 10} + \frac{4}{- 10}$

Divide $a$ by $1$ .

$a = \frac{\ln (45)}{- 10} + \frac{4}{- 10}$

Simplify the right side.

Simplify each term.

Move the negative in front of the fraction.

$a = - \frac{\ln (45)}{10} + \frac{4}{- 10}$

Cancel the common factor of $4$ and $- 10$ .

Factor $2$ out of $4$ .

$a = - \frac{\ln (45)}{10} + \frac{2 (2)}{- 10}$

Cancel the common factors.

Factor $2$ out of $- 10$ .

$a = - \frac{\ln (45)}{10} + \frac{2 \cdot 2}{2 \cdot - 5}$

Cancel the common factor.

$a = - \frac{\ln (45)}{10} + \frac{2 \cdot 2}{2 \cdot - 5}$

Rewrite the expression.

$a = - \frac{\ln (45)}{10} + \frac{2}{- 5}$

Move the negative in front of the fraction.

$a = - \frac{\ln (45)}{10} - \frac{2}{5}$

The result can be shown in multiple forms.

Exact Form:

$a = - \frac{\ln (45)}{10} - \frac{2}{5}$

Decimal Form:

$a = - 0.78066624 \dots$