Weird functional equation $(e^x-1)f(2x)=\left(e^{2x}-1\right)f(x)$ I had a rehearsal test today and got this question in the test that completely stumped me.

Let $f: \mathbb{R} \to \mathbb{R}$ be a non constant continuous function such that
$$(e^x-1)f(2x)=\left(e^{2x}-1\right)f(x)$$
If $f'(0)=1$, then what are $f(x)$ and $f(2x)$?

My try:
I just differentiated both sides with respect to $x$ and made use of $f'(0)=1$, but that got me nowhere.
Please help me someone. Thanks in advance.
 A: I also attended the same test. Here's how I solved it . Its easier than you think it is.
First rearrange both sides
$$\frac{f(2x)}{(e^{2x}-1)}= \frac{f(x)}{(e^{x}-1)}$$
From this you can conclude that $f(x)= e^{x}-1$,[NOTE: Not necessarily $e^x-1$. Please refer Daniel Fischers comment below.]
Otherwise the above condition would cease to be true. Also, $f'(0) = 1$ holds true for the above function.
As it is an objective question paper this is the quickest (and apparently the only single method). Now you can solve for the options given in the question paper.
Note: This question was an MCQ in which multiple options are correct.
A: Note that if $x\ne 0$, we have
$$\frac{f(2x)}{e^{2x}-1}=\frac{f(x)}{e^x-1}$$
So put $g(x)=\dfrac{f(x)}{e^x-1}$ for $x\ne 0$. Then we have
$$g(2x)=g(x)$$
for all non-zero $x$. $g$ is continuous on $\mathbb R-\{0\}$, so we must have
$$g(x) = \begin{cases}
c_+,  & \text{if $x > 0$} \\
c_-, & \text{if $x < 0$}
\end{cases}$$
for some constants $c_+,c_-$.
(Edited to add: It's not that simple $-$ see Daniel Fischer's comment below.)
This gives us
$$f(x) = \begin{cases}
c_+(e^x-1),  & \text{if $x > 0$} \\
c_-(e^x-1), & \text{if $x < 0$}
\end{cases}$$
By continuity of $f$, $c_+=c_-=c,$ say. So $f(x)=c(e^x-1)$ for all $x$.
I will leave the rest to you.
