Explain: How is the value of activation energy determined based on the Arrhenius equation?

(N/A) Method-$1$: Calculation of $E_{a}$:
The Arrhenius equation is $k = A \cdot e^{-\frac{E_{a}}{RT}}$.
Taking the natural logarithm on both sides gives $\ln k = -\frac{E_{a}}{RT} + \ln A$.
If the rate constants at temperatures $T_{1}$ and $T_{2}$ are $k_{1}$ and $k_{2}$ respectively,we have:
$\ln k_{1} = -\frac{E_{a}}{RT_{1}} + \ln A$ $(i)$
$\ln k_{2} = -\frac{E_{a}}{RT_{2}} + \ln A$ $(ii)$
Subtracting $(i)$ from $(ii)$ gives $\ln \frac{k_{2}}{k_{1}} = \frac{E_{a}}{R} (\frac{1}{T_{1}} - \frac{1}{T_{2}})$.
Converting to base-$10$ logarithm: $\log \frac{k_{2}}{k_{1}} = \frac{E_{a}}{2.303 R} (\frac{T_{2} - T_{1}}{T_{1} T_{2}})$.
By substituting the known values of $k_{1}, k_{2}, T_{1}, T_{2}$ and the gas constant $R$,$E_{a}$ can be calculated.
Method-$2$: Graphical Method:
Plot $\ln k$ versus $\frac{1}{T}$ or $\log k$ versus $\frac{1}{T}$.
The plot is a straight line with a slope equal to $-\frac{E_{a}}{R}$ (for $\ln k$) or $-\frac{E_{a}}{2.303 R}$ (for $\log k$).
Thus,$E_{a} = -\text{slope} \times R$ or $E_{a} = -\text{slope} \times 2.303 R$.