Question 12

If $$\alpha, \beta$$ are the values of $$x$$ satisfying the equation $$3\sqrt{\log_2 x} - \log_2 8x + 1 = 0$$, where $$\alpha < \beta$$, then the value of $$\left(\frac{\beta}{\alpha}\right)$$ is

Let $$t = \sqrt{\log_2 x}$$

So $$\log_2 x = t^2$$ and $$\log_2 8x = 3+t^2$$.

Now, the equation becomes $$3t - (3+t^2) + 1 = 0$$

$$t^2-3t+2=0$$

$$t=1$$ or $$t=2$$

So $$x = 2$$ or $$x = 16$$.

Thus $$\alpha = 2$$, $$\beta = 16$$, and $$\dfrac{\beta}{\alpha} = 8$$.

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