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I'm currently studying exponential functions, however I've come across this before and it was just a law that I was told to accept without much proof or explanation, could someone explain the reasoning behind this please. Thank you.

Edit: I am referring to the Law of exponents that states this rule

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How do you define $x^{\frac12}$? As $\exp(\frac12\ln x)$? –  Hagen von Eitzen Oct 23 '13 at 20:28
You would have to tell us how you define $x^{\frac12}$, but you will probably have $\left(x^{\frac12}\right)^2=x^1=x$. –  Carsten Schultz Oct 23 '13 at 20:29
Here is a wonderful explanation of what you're looking for.. –  costashatz Oct 23 '13 at 20:29
Hmm I think I tagged this wrong, it's more to do with the law of exponents, but as I was studying exponential functions this popped into my head. –  seeker Oct 23 '13 at 20:29
The "exponential function" referred to by the tag is the function I mentioned in my answer that takes $x$ to $e^x$. It's not really appropriate for your question. –  MJD Oct 23 '13 at 21:03

2 Answers 2

up vote 65 down vote accepted

When $m$ and $n$ are integers, we have the important law that $$x^m\cdot x^n =x^{m+n}$$

We'd like this law to continue to hold when we define $x^\alpha$ for fractional $\alpha$, unless there's a good reason it shouldn't. If we do want it to continue to hold for fractional exponents, then whatever we decide that $x^{1/2}$ should mean, it should obey the same law: $$x^{1/2}\cdot x^{1/2} = x^{1/2+1/2} = x^1 = x$$

and so $x^{1/2} = \sqrt x$ is the only choice.

Similarly, what should $x^0$ mean? If we want the law to continue to hold, we need $$x^0\cdot x^n = x^{0+n} = x^n$$ and thus the only consistent choice is $x^0 = 1$. And again, why does $x^{-1} = \frac1x$? Because that's again the only choice that preserves the multiplication law, since we need $x^{-1}\cdot x^{1} = x^{-1+1} = x^0 = 1$.

But there is more to it than that. Further mathematical developments, which you may not have 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.

But we can see just by inspection that taking $x=0$ in this formula gives $e^0 = 1$ because all the terms vanish except the first. And similarly, if we put in $x=\frac12$ we get approximately $1.648$, which is in fact the value of $\sqrt e$.

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.

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Excellent explanation, thank you! –  seeker Oct 23 '13 at 20:38
Can I give you some more points somehow? –  seeker Oct 23 '13 at 20:52
Your thanks are my reward. –  MJD Oct 23 '13 at 21:00
@Assad: One of the choices when starting a bounty is "reward a good answer." You must wait 2 days before you can start a bounty, however. –  BlueRaja - Danny Pflughoeft Oct 23 '13 at 22:33
We might think it's equally sensible to define the special case $0^0$ as $0$, but the mathematical convention for that is indeed $0^0 = 1$ in accord with the explanation above. –  minopret Nov 9 '13 at 22:55

For $x\geq 0 $ and by definition $\sqrt x$ is the positive real $y$ such that $y^2=x$ and since $$\left(x^p\right)^q=x^{pq}$$ then for $p=\frac1 2$ and $q=2$ we see that $x^{1/2}=\sqrt x$

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