8. In a chemical reaction two substances combine to form a third substance. At time \(t , t \geq 0\), the concentration of this third substance is \(x\) and the reaction is modelled by the differential equation $$\frac { \mathrm { d } x } { \mathrm {~d} t } = k ( 1 - 2 x ) ( 1 - 4 x ) \text {, where } k \text { is a positive constant. }$$

1. Solve this differential equation and hence show that $$\ln \left| \frac { 1 - 2 x } { 1 - 4 x } \right| = 2 k t + c \text {, where } c \text { is an arbitrary constant. }$$
2. Given that \(x = 0\) when \(t = 0\), find an expression for \(x\) in terms of \(k\) and \(t\).
3. Find the limiting value of the concentration \(x\) as \(t\) becomes very large. END