Why do negative exponents work the way they do? [closed]
Why is a value with a negative exponent equal to the multiplicative inverse but with a positive exponent?
$$a^{-b} = \frac{1}{a^b}$$
If you know that $$x^y \cdot x^z = x^{y+z} \tag{1}$$ and that $$x^0 = 1 \tag{2}$$ then $$\begin{align} 1 &= 1 \\ a^0 &= 1 \\ a^{b - b} &= 1\\ a^{b} \cdot a^{-b} &= 1 \\ a^{-b} &= \dfrac{1}{a^b}\end{align}$$
When $m$ and $n$ are integers, we have the important law that $$x^m\cdot x^n =x^{m+n}$$
The reason for this is easy to see: $x^3$ means $x\cdot x \cdot x$, and $x^2$ means $x\cdot x$, so $$\begin{align}x^3 \cdot x^2 &= (x\cdot x \cdot x)\cdot(x\cdot x)\\ &=x\cdot x \cdot x\cdot x \cdot x \\& = x^5 \end{align}$$
We'd like this law to continue to hold when we define $x^\alpha$ for negative $\alpha$, unless there's a good reason it shouldn't. If we do want it to continue to hold for negative exponents, then whatever we decide that $x^{-1}$ should mean, it should obey the same law: $$x^{-1}\cdot x^{2} = x^{-1+2} = x^1 = x$$
and so $x^{-1} = \frac1x$ is the only choice.
Similarly, what should $x^{-3}$ mean? If we want the law to continue to hold, we need $$x^{-3}\cdot x^{3} = x^{-3+3} = x^0 = 1$$ and thus the only consistent choice is $x^{-3} = \frac1{x^3}$.
But there is more to it than that. Further mathematical developments, which you have not seen yet, confirm these choices. For example, one shows in analysis that as one adds more and more terms of the infinite sum $$1 + x + \frac{x^2}2 + \frac{x^3}6 + \frac{x^4}{24} + \cdots$$ the sum more and more closely approaches the value $e^x$, where $e$ is a certain important constant, approximately $2.71828$. One can easily check numerically that this holds for various integer values of $x$. For example, when $x=1$, and taking only the first five terms, we get $$1 + 1 + \frac12 + \frac16 + \frac1{24}$$
which is already $2.708$, quite close to $e^1$, and the remaining terms make up the difference. One can calculate $e^2$ by this method and also by straightforward multiplication of $2.71828\cdot2.71828$ and get the same answer.
If you put $x=-1$ in this formula, you get $$e^{-1} \stackrel?= 1 -1 +\frac12 - \frac16 + \frac1{24}\cdots$$ and adding up just the first few terms one gets $0.375$, which is already pretty close to $\frac1e \approx 0.368$.
If it didn't work out this way, we would suspect that something was wrong somewhere. And in fact it has often happened that mathematicians have tried defining something one way, and then later developments revealed that the definition was not the right one, and it had to be revised. Here, though, that did not happen.
(Much of this is copied from my answer to a similar question earlier.)