Logaritmic equation

Question

Solve the system of equations

$20=x(1-e^{-k})$

$30=x(1-e^{-2k})$

Can anybody explain how to find $k$ and $x$ from this 2 equations?

Answer 1

Divide one by another one, then cross-multiply and you get
$2(1-e^{-2k})=3(1-e^{-k})$. That is a quadratic equation.

Answer 2

Hint: $\displaystyle\frac32=\frac{x(1-e^{-2k})}{x(1-e^{-k})}=\frac{1-(e^{-k})^2}{1-e^{-k}}$.