In the graph drawn between $\log \mathrm{K}$ and $1/\mathrm{T}$, intercept equals to:

The Arrhenius equation is given as:

                $\mathrm{logk}=\mathrm{logA}-\frac{{\mathrm{E}}_{\mathrm{a}}}{2.303\mathrm{RT}}$

This equation can be compared with the straight line equation:

              $\mathrm{y} = \mathrm{c} + \mathrm{mx}$

As the slope in the straight line equation is c, and then the slope in the Arrhenius equation will be log A.