Chemical reaction modeled by a differential equation I am badly stuck on the following question. Thus, I am asking for some help :)

Consider a chemical reaction in which compounds $A$ and $B$ combine to form a third compound $X$. The reaction can be written as
$$A + B \xrightarrow{k} X$$ 
If $2 \, \rm{g}$ of $A$ and $1 \, \rm{g}$ of $B$ are required to produce $3g$ of compound $X$, then the amount of compound $x$ at time $t$ satisfies the differential equation
$$ \frac{dx}{dt} = k \left(a - \frac{2}{3}x\right)\left(b - \frac{1}{3}x\right)  $$
where $a$ and $b$ are the amounts of $A$ and $B$ at time $0$ (respectively), and initially none of compound $X$ is present (so $x(0) = 0$). Time is in units of minutes, and $k$ is the reaction rate, per minute per gram.
Use separation of variables (and integration by partial fractions) to show that solution can be expressed in the form
$$  \ln \left(\frac{a - \frac{2}{3}x(t)}{b - \frac{1}{3}x(t)} \cdot \frac{b}{a}\right) = ct  $$
where the constant c depends on k,a,b. Now suppose that a = 15g, b = 20g and after 10min, 15g of compound X has been formed, find the amount of X after 20 mins.

So, first of all, showing this solution can be expressed as second equation,
by using separation of variables ( and integration by partial fractions), i got
$$  \ln \left(\frac{a - \frac{2}{3}x}{b - \frac{1}{3}x}\right) * \frac{3}{a-2b} = kt + c $$
how can this can be expressed in the form of the second equation ? am I doing something wrong ?, I dont get how I should get  $$ \frac{b}{a} $$ inside the ln... and on the RHS, how kt + c becomes just ct ?
and by the second euqation, 
"""
where the constant $c$ depends on $k$, $a$ and $b$. Now suppose that a = 15g, b = 20g and after 10min, 15g of compound X has been formed, find the amount of X after 20 mins.
"""
how do we find the amount of $X$ after $20$ minutes?
Thank you !
 A: Without checking if the two forms are both correct :
$$\begin{cases}
\ln \left(\frac{a - \frac{2}{3}x(t)}{b - \frac{1}{3}x(t)} \cdot \frac{b}{a}\right) = ct  \qquad [1]\\
\ln \left(\frac{a - \frac{2}{3}x}{b - \frac{1}{3}x}\right) * \frac{3}{a-2b} = kt + c \qquad [2]
\end{cases}$$
The relationship between $a,b,c,k$ can be derived :
$\begin{cases}
\ln \left(\frac{a - \frac{2}{3}x(t)}{b - \frac{1}{3}x(t)} \cdot \frac{b}{a}\right) =\ln \left(\frac{a - \frac{2}{3}x(t)}{b - \frac{1}{3}x(t)}\right) +\ln\left(\frac{b}{a} \right) =ct \\
\ln \left(\frac{a - \frac{2}{3}x(t)}{b - \frac{1}{3}x(t)} \cdot \frac{b}{a}\right) = \frac{a-2b}{3}(kt+c)
\end{cases} \quad\to\quad ct-\ln\left(\frac{b}{a} \right)=\frac{a-2b}{3}(kt+c)$
$$c=\frac{a-2b}{3}k=-\ln\left(\frac{b}{a} \right)$$
If this relationship is true, Eqs. $[1]$ and $[2]$ are equivalent.
$\begin{cases}
a=\frac{3c}{k}\frac{e^c}{e^c-1}\\
b=\frac{3c}{k}\frac{1}{e^c-1}\\
\end{cases}$
A: The given ODE can be rewritten in the form
$$\dot x = \frac{2 \kappa}{9} \left( x - \frac{3 a}{2} \right) \left( x - 3 b \right)$$
or, alternatively, in the form
$$\left( \frac{1}{x - \frac{3 a}{2}} - \frac{1}{x - 3 b} \right) \, \mathrm{d} x = \left(\frac{a - 2 b}{3}\right) \kappa \,\mathrm{d} t$$
Integrating,
$$\ln \left[ \left( \frac{x - \frac{3 a}{2}}{x - 3 b} \right) \left( \frac{x_0 - 3 b}{x_0 - \frac{3 a}{2}} \right) \right] = \left(\frac{a - 2 b}{3}\right) \kappa t$$
If $x_0 = 0$, then
$$\ln \left[ \left( \frac{2 x - 3 a}{x - 3 b} \right) \frac{b}{a} \right] = \left(\frac{a - 2 b}{3}\right) \kappa t$$
A: The special case $a=2b$ is missing in the answers.
In this case
$$
x(t)=3b-\frac{9b}{2b\kappa t+3}.
$$
