Functional equation, find particular value given $f(ab)=bf(a)+af(b)$ Let $f(x)$ be a function such that $f(ab) = bf(a) + af(b)$ for all nonzero real numbers. Given that $f(4) = 3$, which of the following is a possible value of $f(2018)$? 
(A) $0\quad$ (B) $\dfrac34\quad$ (C) $\dfrac43\quad$ (D) $1512\quad$ (E) $2688\quad$
By substitution, it's easy to find that $f(1) = 0$ and $f(2) = \dfrac34$. But how can I get to $f(2018)$?

Let $g(x)$ satisfy the Cauchy functional equation $g(x+y)=g(x)+g(y)$. Let $f(x)=x \, g(\log |x|)$.
Then
$$f(ab)=ab \,g(\log |a| + \log |b|)=ab\, g(\log |a|)+ab\,g(\log |b|)=b\,f(a)+a\,f(b)$$
and so $f$ satisfies the functional equation, for any such $g$. If there are no further continuity requirements, we can choose $g$ to have any values we want at non-commensurable $x,y$, as in this answer. Since $\log 4$ and $\log 2018$ are non-commensurable, that means we can choose any value we want for $f(2018)$.
On the other hand, with any continuity (or even measurability) requirements at all, $g(x)=kx$. Since $f(4)=3$, $g(\log 4)=\frac{3}{4}$, and so $g(x)=\frac{3}{4\log 4}x$.
In this case, it follows that $f(2018)=2018 \left(\frac{3}{4} \log_4 2018\right)=\frac{3027}{2} \log_4 2018$, which is not one of the answers given.
