How many (positive real) zeroes can $f(x)=x^a-b(c^x)$ have?

Question

Let $a,b,c$ be positive real numbers. Define $f:(0,\infty) \to \mathbb R$ by $f(x)=x^a-b(c^x)$. What is the maximal numbers of zeroes that $f$ can have ? My guess is that the answer is $2$ (note that if $b=\frac{1}{2^a}$ and $c=2^a$, then $1$ and $2$ are zeroes of $f$).

Answer 1

The problem does not change if we replace $(a,b,c)$ with $(1,b^{\frac{1}{a}},c^{\frac{1}{a}})$, so we may assume that $a=1$.

Suppose that $f$ has at least three zeroes, $x_1<x_2<x_3$. Then $b=\frac{x_2}{c^x_2}=\frac{x_3}{c^x_3}$, so that $c=\bigg(\frac{x_2^{x_3}}{x_3^{x_2}}\bigg)^{\frac{1}{x_3-x_2}}$ and hence $c>1$. The map $g(x)=c^x$ is convex on $(0,\infty)$, so we must have for any $x>0$,

$$ g(x) \geq g(x_2)+g'(x_2)(x-x_2) $$

or in other words, $c^x \geq c^{x_2}+\ln(c)c^{x_2}(x-x_2)$ so that $b(c^x) \geq x_2-\ln(c)x_2(x-x_2)$ and $b(c^x)-x \geq (x_2-x)(1+\ln(c)x_2)$. Taking $x=x_1$, we deduce $0 \geq (x_2-x_1)(1+\ln(c)x_2)$ which is impossible because $c>1$ and $x_2>0$.