I was asked this question and got completely stumped. The person who asked this has instructed me specifically not to use logarithms. I know there is a more sophisticated way to explain exponentiation using limits of sequences (which would work here), but unfortunately limits are outside the reach in this particular context. So I was wondering, is there a simple way to explain why $0<x<1$ can only be expressed as negative powers of $10$ (or any other number greater than $1$, for that matter) using only high-school algebra? I suppose the proof doesn't have to be entirely rigorous, as long as it's sufficiently convincing.
How to prove $10^x> 1$ for $x>0$ without using logarithms
algebra-precalculus
Related Solutions
We need to bring out a non-elementary (termed "special") function for this kind of problem; the one intended for this type of situation is the Lambert W function. It is defined to be the inverse function of $xe^x$ - that is, it satisfies the functional equation $W(x)e^{W(x)}=x$.
Now we wish to solve $ax=a^x$. Fix $a$, so that we're solving for $x$ in terms of $a$. Rearrange:
$$ax=e^{(\ln a)x}\iff xe^{-(\ln a)x}=a^{-1}\iff \color{Blue}{-(\ln a)x} e^{\color{Blue}{-(\ln a)x}}=\color{DarkGreen}{-(\ln a)/a}$$
$$\iff \color{Blue}{-(\ln a)x}=W\left(\color{DarkGreen}{-\frac{\ln a}{a}}\right)\iff x=-\frac{1}{\ln a}W\left(-\frac{\ln a}{a}\right).$$
Inverse functions can trip a lot of people up when they're new to the concept, and in particular, those that are inverse functions which are non-elementary (not able to be expressed with a finite number of the basic four operations combined with roots and powers and exponentials) can be pretty trick and take a bit of getting used to. The Lambert W function falls into this category occasionally, which is why I provide a full solution as an introduction on how to manipulate it.
Here is my second try. I give some references for Olympiad-style problem solving. Hopefully you'll find something useful in each of them.
Topics in Algebra and Analysis: Preparing for the Mathematical Olympiad by Bulajich, Gómez and Valdez is what most closely resembles a comprehensive treatment among the books I know. Very student-friendly.
The Art and Craft of Problem-Solving teaches basic-level problem solving, including an algebra section.
Problem-solving strategies by Engel is a famous compendium of problems. The focus being on effective problem-solving, theory is really scant but perusing the algebra sections you may find interesting problems.
101 Algebra Problems from the Training of the USA IMO Team by Andreescu and Feng is a more specialized compendium.
Putnam and Beyond by Gelca and Andreescu is a textbook focusing on undergraduate-level contests. Here you can find many challenging problems from areas usually excluded from high school contests (e.g. calculus and linear algebra).
Polynomials by Barbeau is a more leisurely treatment of the basic theory of polynomials (in case you're dissatisfied with any of the previous suggestions).
Complex Numbers from A to Z by Andreescu and Andrica is a comprehensive exposition about complex numbers. If you do have to teach this topic, I strongly recommend you to take a look.
As a final warning, I must tell you that (at least in my own experience) contest-geared textbooks tend to focus in quick development of problem-solving skills rather than rigorous mathematical exposition. You may want to consider other kind of textbook to compensate for this.
Best Answer
If $x$ is a positive rational number $p/q$ (where $p$ and $q$ have no common factors), then we can define $10^x$ as $$ \left(\sqrt[q]{10}\right)^p=\underbrace{\sqrt[q]{10} \cdot \sqrt[q]{10} \cdot \sqrt[q]{10} \cdot \ldots \cdot \sqrt[q]{10}}_{\text{$p$ times}} \, . $$ For any positive integer $q$, the $q$-th root of $10$ is greater than $1$. This means that $10^x$ is greater than $1$ because the product of $p$ numbers that are greater than $1$ is greater than $1$.
There are no hiccups caused when $x$ is irrational, because in that case $10^x$ is defined in such a way that it is strictly increasing.