Exponential functions!

Ok, this problem literally took me so long! But here it goes:

Let $m = \ln(2)$ and $n = \ln(3) \implies 2 = e^m$ and $3 = e^n$

$\implies f(x) = e^{mx}$ and $g(x) = e^{nx} \implies f'(a) = me^{ma} = g'(b) = ne^{nb} \implies$ $a = \dfrac{1}{m}\ln(\dfrac{n}{m}e^{nb}) =\dfrac{1}{m}\ln(\dfrac{n}{m}) + \dfrac{n}{m}b$

$\implies$

$A:(\dfrac{1}{m}\ln(\dfrac{n}{m}) + \dfrac{n}{m}b,\dfrac{n}{m} * e^{nb})$ and $B:(b,e^{nb})$

$\implies m_{AB} = \dfrac{(n - m)e^{nb}}{\ln(\dfrac{n}{m}) + (n - m)b} = ne^{nb} \implies$ $b = \dfrac{n - m - n\ln(\dfrac{n}{m})}{n(n - m)}$

$\implies m_{AB} = g'(b) = \dfrac{ne}{(\dfrac{n}{m})^{\frac{n}{n - m}}} = (\dfrac{m^n}{n^m})^{\frac{1}{n - m}} * e$

Replacing $m = \ln(2)$ and $n = \ln(3)$ into $m_{AB}$ we obtain:

$m_{AB} = (\dfrac{(\ln(2))^{\ln(3)}}{(\ln(3))^{\ln(2)}})^{\dfrac{1}{\ln(\frac{3}{2})}} * e \approx \boxed{0.85740343}$