Prove that $m=n$ in the equation $a^{m-n} = 1$ for $a \gt 1$

Question

If you have the equation $a^{m-n} = 1$, how would you prove that $m=n$, with the constraint of $a \gt 1$? Assuming $a, m$, and $n$ are all real numbers.

To explain what I'm talking about, I'm only going to test this out. You would select a valid value for $a$. In this case, I will select $2$. Then, plug in values of $m$ and $n$ when $m=n$, and see if the equation is true. Then, a counter-example, when $m \neq n$.

$$2^{m-n} = 1\Rightarrow 2^{3-3} = 1\Rightarrow2^0=1\Rightarrow1=1$$ The equation is true when $m=n$.

$$2^{m-n} = 1\Rightarrow 2^{4-3} = 1\Rightarrow2^1=1\Rightarrow2=1$$ The equation is not true when $m \neq n$, so $2$ is a valid value.

Answer 1

Suppose that $$ a^{m - n} = 1, \ \ \mbox{where} \ a > 1 $$

Taking logarithm on both sides, we get $$ (m - n) \ \ln a = 0 $$ which gives $$ m - n = 0 $$ or equivalently $$ m = n $$

Answer 2

We can assume that : $$a^0=1,a>1$$

Hence : $$a^{m-n}=a^0=1,a>1,m\in \mathbb{R},n\in \mathbb{R}$$

By default : $$a^{m-n}=a^0,a>1,m\in \mathbb{R},n\in \mathbb{R}$$

Thus, we get : $$m-n=0,m\in \mathbb{R},n\in \mathbb{R}$$

Therefore: $$m=n,m\in \mathbb{R},n\in \mathbb{R}$$