Proof that $a^x$ < $a^{x + \Delta x}$

Question

How can I prove that $a^x$ < $a^{x + \Delta x}$, where $\Delta x > 0$, $a$ is a constant, and $a > 1$ (in my case, $a=2$)? I don't want to use a graph, of course.

Answer 1

It all depends on what you know. Do you know that if $a\gt 1$ and $b\gt 0$ then $a^b\gt 1$? If so, $2^{x+\Delta x}-2^x = 2^x(2^{\Delta x}-1)$ is a product of positive numbers.