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Can you use noninteger powers Like is $x^{8.3} / x^{2.2} = x^{6.1}$?

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6  
"dora want to know" is a lousy title. Look around the site and see that titles are supposed to be informative. –  Arturo Magidin Mar 18 '11 at 13:32
    
Short answer: Yes. –  Brian Mar 18 '11 at 13:36
    
Yes, and if you rewrite, say, $x^{8.3}$ as $x^{\frac{83}{10}}$, you can represent the variable as $(\sqrt[10]{x})^{83}$, by the property of the exponents. Same with the other variables inyour question. –  InterestedGuest Mar 18 '11 at 13:40
    
As others have said, yes, but the base must be nonnegative ($x\ge 0$) for things to be well-defined (this is covered in Arturo's answer, but is really critical; his end note about extending to negative bases with complex numbers is true, but has some caveats about uniqueness and choices of principal values). –  Isaac Mar 18 '11 at 18:36

2 Answers 2

Yes, you can. The definitions grow every more complicated depending on what kinds of exponents you allow, but the laws of exponents hold in general.

For positive integer powers, you can define exponentials ("powers") for any base $x$: \begin{align*} x^1& = x\\ x^{n+1} &= x^nx \end{align*} which is equivalent to $$x^n = \underbrace{x\times x\times\cdots\times x}_{n\text{ factors}}.$$

This satisfies the usual "exponent rules": $x^nx^m = x^{n+m}$, and $(xy)^n = x^ny^n$.

If you want to extend it to exponent $0$, you now have to exclude $x=0$ ($0^0$ is usually left undefined in general; see this previous question). You define $x^0 = 1$ for all $x\neq 0$.

Then you can also extend to negative integer exponents: if $n\gt 0$, then we define $x^{-n} = \frac{1}{x^n}$. This adds the third "exponent rule": $x^n/x^m = x^{n-m}$.

You can then extend the exponents to fractions, but now you have to restrict the base to positive real numbers in general. If $p$ is an integer, we define $x^{1/p} = \sqrt[p]{x}$, the $p$th root of $x$ (the unique positive real number $y$ such that $y^p= x$.

Then, if $p$ and $q$ are integers, with $q\gt 0$, we define $x^{p/q} = (x^p)^{1/q} = \sqrt[q]{x^p}$.

With this definition, the exponentiation rules still hold; this applies to the case you have above, since $8.3 = \frac{83}{10}$, $2.2 = \frac{22}{10}$, and $6.1 = \frac{61}{10}$.

You can extend the definition further, to any real number as an exponent, through the use of either limits or logarithms. We define $a^b = e^{b\ln(a)}$, provided $a\gt 0$. This still satisfies the exponent laws.

(You can extend the definition even further by allowing complex numbers as answers; then you can remove the restriction on the base being positive).

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Yes you can. In fact, $$\frac{1}{x^y}=x^{-y}$$ for $x,y \in \mathbb{R}, x \neq 0$. So $$\frac{x^y}{x^z}=x^{y-z}$$ for $x,y,z \in \mathbb{R}, x \neq 0$.

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