China Mathematical Competition

By the condition, it follows that $\cos a + \log_2 a = \cos 2a + \log_2 2a = 2\cos^2 a - 1 + 1 + \log_2 a$, so $\cos a = 2\cos^2 a$. Thus, we have $\cos a = 0$ or $\cos a = \frac{1}{2}$, and hence correspondingly $\cos 2a = 2\cos^2 a - 1 = -1$ or $\cos 2a = -\frac{1}{2}$. Therefore,
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$$\text{f(2a) - f(4a) = 2a + 2 2a - 4a - 2 4a = 2a - 2 2 2a = -3, if 2a = -1, -1, if 2a = - 1 2 .}$$
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