Second order reaction with different reactants In a laser photolysis reaction $\ce{C6H5NH2}$ $(C_0 = \pu{60 µM})$ is oxidized to its radical cation. The disappearance of the cation can follow three pathways: (1) Dimerzing of the cation, (2) Reaction between the cation of the initial substance, (3) Reaction between the cation and the solvent. Decide which pathway is most likely to occur.
$$
\begin{array}{l|cccccccc}
\hline
t/\pu{µs}~(\text{efter puls}) &amp; 0 &amp; 5 &amp; 10 &amp; 15 &amp; 20 &amp; 25 &amp; 30 &amp; 35 \\
\hline
[\text{Radikalkatjon}]/\pu{µM} &amp; 20 &amp; 11.12 &amp; 6.86 &amp; 4.47 &amp; 3.00 &amp; 2.06 &amp; 1.43 &amp; 1 \\
\hline
\end{array}
$$

I first checked for the dimerizing of the cation by assuming it to be a second order reaction with the same reactants and plotting $1/[\mathrm{cation}]$ against $t,$ which didn't give a good fit. So I wanted to plot it as a second order reaction with different reactants $(\ce{A + B -&gt; products})$ but I don't understand how that is done.
I should plot $\displaystyle\ln\frac{a-x}{b-x}$ against $t,$ but what exactly is $(a-x)$ and $(b-x)?$
I know that I should set up a reaction like:
$$
\begin{array}{l|ccc}
\hline
&amp; \text{A} &amp; \text{B} &amp; \text{product} \\
\hline
t=0 &amp; a &amp; b &amp; 0 \\
t=t &amp; a-x &amp; b-x &amp; x \\
\hline
\end{array}
$$
And I would assume that my $a = \pu{20 µM}$ and $b = \pu{60 µM}.$ But where do I go from there? All help is appreciated!
The general solution, if the reaction is $aA+bB \to product$, is to allow an amount $x$ to react at time $t$ then there will be concentrations $(A_0-ax)$ and $(B_0-bx)$ of species A and B respectively at time $t$ if $A_0,B_0$ are the concentrations at time zero.  The rate equation is $\displaystyle  -\frac{1}{a}\frac{d(A_0-ax)}{dt}=\frac{dx}{dt}=k(A_0-ax)(B_0-bx)$ which can be integrated ($0\to x, 0\to t)$ to give

$$ \ln\left(\frac{A_0-ax}{B_0-bx} \right)=k(A_0b-B_0a)t +\ln\left(\frac{A_0}{B_0} \right)$$

If the reaction follows this scheme plot the left hand side vs $t$ and this should produce a straight line of slope $k(A_0b-B_0a)$.

Should it happen that concentrations are initially $A_0/B_0 = a/b$ so that species are present in stoichiometric amounts then as $A_0b-B_0a=0$ the  solution above cannot be used. In this case the rate equation becomes $\displaystyle  \frac{dA}{dt}=-bkA^2$ which gives $\displaystyle \frac{1}{A}-\frac{1}{A_0}=bkt$.