Finding the root of the problem

$\begin{aligned} \frac {40}{x-1} - \frac {160}{x-4} - \frac {200}{x-5} + \frac {320}{x-8} & = 6x^2 - 27x \\ \frac {40}{x-1} - \frac {200}{x-5} + \frac {320}{x-8} - \frac {160}{x-4} & = 3x(2x - 9) \\ \frac {40x-200 - 200x+200}{(x-1)(x-5)} + \frac {320x-1280 - 160x+1280}{(x-8)(x-4)} & = 3x(2x - 9) \\ - \frac {160x}{(x-1)(x-5)} + \frac {160x}{(x-8)(x-4)} & = 3x(2x - 9) \\ \frac {160x(x^2-6x+5-x^2+12x-32)}{(x-1)(x-4)(x-5)(x-8)} & = 3x(2x - 9) \\ \frac {160x(6x-27)}{(x-1)(x-4)(x-5)(x-8)} & = 3x(2x - 9) \\ 3x(2x-9) \left(\frac {160}{(x-1)(x-4)(x-5)(x-8)} -1 \right) & = 0 \end{aligned}$

Therefore, there are three real roots to the equation $x=0$ , $x = \dfrac 92$ and:

$\begin{aligned} \frac {160}{(x-1)(x-4)(x-5)(x-8)} & = 1 \\ \implies (x-1)(x-4)(x-5)(x-8) & = 160 \end{aligned}$

We note that when $x=0$ , $(x-1)(x-4)(x-5)(x-8) = (-1)(-4)(-5)(-8) = 160$ , so is when $x = 9$ , $(9-1)(9-4)(9-5)(9-8) = (8)(5)(4)(1) = 160$ . Since $x=0$ is already a root the third root is $x=9$ .

Putting $\alpha = 9$ and $\beta = \dfrac 92$ , then $\log_\alpha (2\beta) = \log_9 9 = \boxed{1}$ .