A proof for If $a>0$ AND $r<s$ then: $a>1⇔a^r<a^s$ without using the natural logarithm function?

Question

We have $a$ is a real positive number different than zero.

$r$ and $s$ belong both to $\mathbb{Q}$

$(r<s)$ AND $(a^r<a^s)$ ⇔ $(r<s)$ AND $(ln(a^r)<ln(a^s))$ ⇔ $(r<s)$ AND $(r.ln(a)<s.ln(a))$

On the other hand we have:

$a>1⇔ln(a)>ln(1)⇔ln(a)>0$

which means that:

r.$ln(a)<s.ln(a) ⇔ r<s$ because $ln(a)>0$

This can be summarized as follows:

If $a>0$ AND $r<s$ then:

$a>1⇔a^r<a^s$

Is there any other proof for: If $a>0$ AND $r<s$ then: $a>1⇔a^r<a^s$ without using the natural logarithm function $ln$?

This was my answer for someone asking for a proof here

Unfortunately he needed a proof without using the $ln$ function.

Answer 1

If $r$ and $s$ are rational numbers and $r<s$, there are integers $k$ and $l$ and a natural number $t$ such that $r=\frac kt$, that $s=\frac lt$, and that $k<l$. So$$a^r=a^{k/t}=\sqrt[t]{a^r}\quad\text{and}\quad a^s=a^{k/t}=\sqrt[t]{a^l}.$$And, since $k<l$, $a^r<a^l$, and therefore $\sqrt[t]{a^r}<\sqrt[t]{a^l}$.

Answer 2

You ask for a proof without logarithmic function. However, we must know if $r$ is real or not.

If it is, then we cannot define $a^r$ without logarithm function same is for $a^s$ if $s$ is real too.

EDIT: Since we know that $r$ and $s$ are rational number, instead of using logarithmic function we can just use $f: x \mapsto \sqrt[q]{a^x}$. This function is defined for all $x, q$ natural integers.

So this simply equivalent to prove that $f$ parametric $a$ is increasing for parameter $a > 1$. This can be done recursively by proving the ratio of two consecutive terms is greater than 1 if and only if $a>1$.