Log Mania

Let $m = \ln(\alpha)$ and $n = \ln(\beta) \implies \alpha = e^m$ and $\beta = e^n$ .

$x > 0 \implies \large f(x) = \log_{e^m}(x) = \dfrac{\ln(x)}{m}$ and $\large g(x) = \log_{e^n}(x) = \dfrac{\ln(x)}{n}$

$f'(a) = \dfrac{1}{ma} = g'(b) = \dfrac{1}{nb} \implies a = \dfrac{n}{m}b \implies$

$A:(\dfrac{n}{m}b, \ln(\dfrac{n}{m}b))$ and $B:(b, \dfrac{\ln(b)}{n}) \implies$

$\large m_{AB} = \dfrac{\ln((\dfrac{n}{m})^n b^{n - m})}{bn(n - m)} = \dfrac{1}{bn} \implies$

$\large \ln((\dfrac{n}{m})^n b^{n - m}) = n - m \implies (\dfrac{n}{m})^n b^{n - m} = e^{n - m}$

$\large \implies \large b = (\dfrac{m}{n})^{\frac{n}{n - m}}e \implies a = (\dfrac{m}{n})^{\frac{m}{n - m}} e$

$\implies$

$\large A:((\dfrac{m}{n})^{\frac{m}{n - m}} e, \ln((\dfrac{m}{n})^{\frac{1}{n - m}} e^{\frac{1}{m}})$

and,

$\large B:((\dfrac{m}{n})^{\frac{n}{n - m}} e, \ln((\dfrac{m}{n})^{\frac{1}{n - m}} e^{\frac{1}{n}})$ .

Using the symmetry about the $y$ axis we have:

$\large A':(-(\dfrac{m}{n})^{\frac{m}{n - m}} e, \ln((\dfrac{m}{n})^{\frac{1}{n - m}} e^{\frac{1}{m}})$

and,

$\large B':(-(\dfrac{m}{n})^{\frac{n}{n - m}} e, \ln((\dfrac{m}{n})^{\frac{1}{n - m}} e^{\frac{1}{n}})$ .

and $\large E:((\dfrac{m}{n})^{\frac{n}{n - m}} e, \ln((\frac{m}{n})^{\frac{1}{n - m}} e^{\frac{1}{m}})$

$\large AA' = 2(\dfrac{m}{n})^{\frac{m}{n - m}}, \:\ BB' = 2(\dfrac{m}{n})^{\frac{n}{m - n}}$ and $\large EB = \dfrac{n - m}{nm} \implies$

$\large A_{AA'B'B} = (\dfrac{m}{n})^{\frac{m}{n - m}} (\dfrac{n + m}{n})(\dfrac{n - m}{nm})$ .

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$\dfrac{\ln(\alpha)}{\ln(\dfrac{\beta}{\alpha})} = 1 \implies \ln(\dfrac{\alpha^2}{\beta}) = 0 \implies \beta = \alpha^2$ and $m = \ln(\alpha)$ and $n = \ln(\beta) = 2\ln(\alpha) \implies A_{AA'BB'} = \dfrac{3e}{8\ln(\alpha)} = \dfrac{e}{16} \implies \alpha = e^6 \implies \beta = e^{12} \implies \lfloor{\alpha\beta}\rfloor = \lfloor{e^{18}}\rfloor = \boxed{65659969}$ .

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Incidentally, the slope $\large m_{AB} = g'(b) = f'(a) = \dfrac{1}{n(\dfrac{m}{n})^{\frac{n}{n - m}} * e} = \dfrac{1}{e}(\dfrac{n^m}{m^n})^{\frac{1}{n - m}} = \dfrac{1}{e}(\dfrac{(\ln(\beta))^{\ln(\alpha)}}{(\ln(\alpha)^{\ln(\beta)}})^{\frac{1}{\ln(\frac{\beta}{\alpha})}}$ .