Does their exist a real continuous function other than $f(x)=0$ such that $f(2x) = -2f(x)$?

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I have a gut feeling it doesn't exist but I'm not sure how to prove/disprove it. My attempt: Suppose there exists $a \in \mathbb{R}\setminus\left\{0\right\}$ such that $f(a) \neq 0$ . Define $x_n = \frac{a}{2^n}$

$f(x_{n+1}) = \frac{-1}{2} f(x_n)$ and inductively $f(x_n) = (\frac{-1}{2})^n f(a)$

What can I do from here?

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(Rewriting achille hui's comment as an answer.)

Yes, there are other functions satisfying that equation. One such function is $f(x) = x \sin(\pi \log_2(x))$.