If we plot graph for the first order reaction involving reactant R, then it will be a straight line if plotted for:

\(log\frac{[R]}{[R]_0}\) against time, t

For a first-order reaction, the rate law is given by:
\[ -\frac{d[R]}{dt} = k[R] \]
Where:
\([R]\) is the concentration of the reactant at time \(t\),
\([R]_0\) is the initial concentration of the reactant,
\(k\) is the rate constant.
The integrated form of the first-order rate law is:
\[ ln\frac{[R]}{[R]_0} = -kt \]
This equation indicates that a plot of \(ln\frac{[R]}{[R]_0}\) against time \(t\) for a first-order reaction will yield a straight line. Therefore, the correct answer is: 3. \(log\frac{[R]}{[R]_0}\) against time, \(t\)