What function or equation can result in higher "division result"? like $\frac{f(x+100)} {f_(x+90)} > \frac{f(x+90)} {f(x+80)}$

I'm trying to make an exp system with division result that increase as the level is higher.

I have tried many polynomial equation, but all of it doesn't do what I want. Simple example:

$$f(x)=x^3$$ $$f(80)=512000$$ $$f(90)=729000$$ $$f(100)=1000000$$ where $$\frac{f(100)}{f(90)} < \frac{f(90)}{f(80)} $$ $$\frac{1000000}{729000} < \frac{729000}{512000} $$ $$1.371742 < 1.423828 $$

I can't figure how to how to achieve $$\frac{g(100)}{g(90)} > \frac{g(90)}{g(80)} $$

Let us take a more general setting: you desire to have a function $g$ that, under condition

$$\begin{cases}a > c\\ \text{with} \ \ b:=\frac{a+b}{2}\end{cases}\tag{1}$$

verifies:

$$\frac{g(a)}{g(b)} > \frac{g(b)}{g(c)}$$

If we consider that $g$ takes $>0$ values, this is equivalent to :

$$g(a)g(c)>g(b)^2$$

$ln$ function being increasing, this is equivalent (under condition (1)) to :

$$\frac12 (\ln(g(a))+\ln(g(c)))> \ln(g(\frac{a+b}{2}))$$

Which means that function $f$ defined by

$$\ln(g(x))=:f(x)\tag{2}$$ must be strictly convex.

Otherwise said, the general solution to your issue is:

$$g(x)=\exp(f(x))$$

for any convex function $f$.

This is the case for example with:

$$g(x):=\exp(x^2)$$