According to the Arrhenius equation, the rate constant $k$ is given by $k = A e^{-E^*/RT}$. Taking the natural logarithm on both sides yields $\ln k = \ln A - \frac{E^*}{RT}$. Converting this to base-10 logarithm gives $\log k = \log A - \frac{E^*}{2.303RT}$. This is the equation of a straight line ($y = mx + c$), where $y = \log k$ and $x = \frac{1}{T}$. Therefore, a plot of $\log k$ versus $\frac{1}{T}$ gives a straight line with a slope of $-\frac{E^*}{2.303R}$. The activation energy ($E^*$) can be calculated from the value of this slope.