Prove $a\cdot 2^x+b\cdot 3^x+c\cdot 7^x$ has at most two real solutions for $a,b,c\ne 0$ 
Prove $a\cdot 2^x+b\cdot 3^x+c\cdot 7^x$ has at most two real solutions.

I can prove a variant with two terms in the following way:
$$a\cdot 2^x+b\cdot 3^x=0$$
$$-\frac{a}{b}=\left(\frac{3}{2}\right)^x$$
If $a,b$ have the same parity there are zero solutions, and if they don't there is one (real) solution. 
But I can't expand that method to more terms. How can I approach this problem?

Edit: Either I missed the title the first time or it was edited later. This note can be skipped.
Note: you would need that at least one of them is nonzero. Otherwise, you have infinitely many roots. Even your "solution" for the case of two terms is missing this. However, the conclusion for the case of two terms is correct if you require one of $a$ or $b$ to be nonzero.
We shall assume that $a$ and $b$ are nonzero. Else, this reduces to a two terms case anyway.

The question is equivalent to solving 
$$a\alpha^x + b\beta^x = -c$$
for $\alpha = 2/7$ and $\beta = 3/7$.

Define $f:\Bbb R \to \Bbb R$ as 
$$f(x) := a\alpha^x + b\beta^x + c.$$
We then have
$$f'(x) = a(\ln\alpha)\alpha^x + b(\ln\beta)\beta^x.$$
Now, since $a \neq 0 \neq b$, we have that $f'$ has at most one real root. (By the two terms case.)
This, in turn, implies that $f$ has at most two real roots. (Use Rolle's theorem to disprove the existence of more than two roots.)

This method can be extended for more number of terms as well!
