Proving that there are $n-1$ roots in $a_1x^{b_1}+a_2x^{b_2}+...+a_nx^{b_n}=0 $ on $(0,\infty)$ 
We know that: 
$a_1,...,a_n\in \mathbb R , \forall a_i\neq0 \\
b_1,...,b_n\in \mathbb R : b_j\neq b_k : \forall j\neq k$
Prove that there are $n-1$ roots in $(0,\infty)$: 
$$a_1x^{b_1}+a_2x^{b_2}+...+a_nx^{b_n}=0  $$

Using induction, for n=1 it's obvious that there are 0 roots.
Suppose the statement is true for n-2 and prove for n-1:
From here on I'm not sure how to continue, I noticed that $x^{b_1}(a_1 + \ldots + a_n x^{b_n - b_1})=0$ but at what point can it come into the induction process ? Also, how does the IVT is applied here ?

Hint:
So now just work with $\;a_1 +a_2x^{b_2-b_1}+\ldots+a_nx^{b_n-b_1}\;$,  differentiate and get less than $\;n\;$ summands. Now use what the other question mentions about the relation between the zeros of the derivative of a function and those of the function itself.
