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I like to understand how things work. It's the only way I'll remember the rule. If anyone has a way of breaking this down into logical steps to get from one to the other... it would be greatly appreciated.

why is

$X^n\cdot X^m = X^{n+m}$ ?

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Is X a real number? matrix? random variable? something else? –  Douglas S. Stones Dec 2 '10 at 11:45

2 Answers 2

up vote 12 down vote accepted

$X^n$ is $X$ multiplied by itself $n$ times. $X^m$ is $X$ multiplied by itself $m$ times. When you multiply these together, you'll get $X$ multiplied by itself $n+m$ times, which is $X^{n+m}$.

Here's an example with $n=2$ and $m=3$: $$X^2 \cdot X^3 = (X \cdot X) \cdot (X \cdot X \cdot X) = X^5.$$

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Thanks that did it for me. –  James T Dec 2 '10 at 5:56

If $n$ and $m$ are positive integers, then remember that $X^n$ is shorthand for $$ \underbrace{X\cdot X\cdots X}_{\text{$n$ factors}}$$

So $X^n\cdot X^m$ is: $$X^n \cdot X^m = \underbrace{(X\cdot X\cdots X)}_{\text{$n$ factors}} \cdot \underbrace{(X\cdot X \cdots X)}_{\text{$m$ factors}} = \underbrace{X \cdot X\cdots X}_{\text{$n+m$ factors}} = X^{n+m}.$$

For negative integers, remember that if $n\gt 0$, then $X^{-n} = \frac{1}{X^n}$. So we have that $$X^{-n}\cdot X^{-m} = \frac{1}{X^n}\cdot \frac{1}{X^m} = \frac{1}{X^n\cdot X^m} = \frac{1}{X^{n+m}} = X^{-(n+m)} = X^{-n-m}.$$

From there, using the other exponent rules, you can also get that it works for exponents that are fractions; and then once you define arbitrary exponents, for any exponent.

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