If $a> 1$ and $x > 0$, prove that $a^x > 1$ without the use of Calculus.

Question

In the study of exponential functions we notice that they are increasing or decreasing. In this way, we have the following question.

If $a> 1$ and $x > 0$, prove that $a^x > 1$ without the use of Calculus.

Answer 1

How about this:

Assume to the contrary that

$ a^x \leq 1 $

Since $x>0$ and $a>1$:

$ \left(a^x\right)^{\frac{1}{x}} \leq (1)^{\frac{1}{x}} \hspace{5 mm} \Rightarrow a \leq 1 $

a contradiction.

Thus, $a^x>1$

Answer 2

Let assume on contrary that $a^x\le 1\;$ Now dividing both sides by a, we get $a^{x-1}\le a^{-1}$ $\implies x-1\le -1 \implies x\le 0$ which is a contradiction. Hence $a^x\gt 1$