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Can someone please explain me, why when we are looking at the function $f(x)=a^x $ , we should remember that $1 \neq a >0 $ ? (And not saying that we can't put an x that satisfies: $ 0 < x < 1 $ ?

Any understandable explanation will be great!

Thanks !

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    Disallowing 0 wouldn't stop any problems . Taking *any* non-integer $x$ would cause problems with negative values of $a$.2012-10-22

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I assume that we are working over the reals.

The general exponential function $a^x$ is usually defined by the formula $a^x=e^{x\ln a}.$ When $a\le 0$, $\ln a$ does not exist.

As for excluding $a=1$, there is no reason to do so, except for the fact that $1^x$ is a very uninteresting function!

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    Congratulations, André, $f$or 100k+. :-)2012-10-22
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Besides to @André's theoretical answer(+1); imagine what would be happen if $a<0$? For example if $a=-4$, what would you do with $(-4)^{\frac{m}{n}}$ wherein $n$ is even and $m$ is odd? Would we have a real number? How many fractions of this type are there in $\mathbb R$? I see the close story can be when $a=1$.