CHEMISTRY HELP!! For equilibrium reaction A <=> B, and from k=Ae^(-Ea/RT), derive equation ln(K1/K2)=deltaH/R(1/T2-1/T1)?

For the equilibrium reaction: A ⇌ B …… Equilibrium constant = K
For the forward reaction, assume that k is the rate constant and Eₐ is the activation energy.
For the backward reaction, assume that k' is the rate constant and Eₐ' is the activation energy.

For the forward reaction: k = A e^(-Eₐ/RT)
For the backward reaction: k' = A' e^(-Eₐ'/RT)

At equilibrium, (forward reaction rate) = (backward reaction rate)
k [A] = k' [B]
[B]/[A] = k/k'

Then, equilibrium constant, K = [B]/[A] = k/k' = [A e^(-Eₐ/RT)]/ [A' e^(-Eₐ'/RT)] = (A/A') e^[-(Eₐ - Eₐ')/RT
But Eₐ - Eₐ' = ΔH, then K = (A/A') e^(-ΔH/RT)
Take logarithm on the both side, then ln(K) = ln[(A/A') e^(-ΔH/RT)]
Thus ln(K) = (-ΔH/RT) + ln(A/A')

When T = T₁, K = K₁: ln(K₁) = (-ΔH/RT₁) + ln(A/A') …… [1]
When T = T₂, K = K₂: ln(K₂) = (-ΔH/RT₂) + ln(A/A') …… [2]

[1] - [2]:
ln(K₁) - ln(K₂) = (-ΔH/RT₁) - (-ΔH/RT₂)
ln(K₁/K₂) = (ΔH/R)(1/T₂) - (ΔH/R)(1/T₁)
Hence, ln(K₁/K₂) = (ΔH/R)[(1/T₂) - (1/T₁)]