In $f(x) = a^x$, I understand why $a > 0$ and $a \neq 1$. But how is $f(x) = a^x$ guaranteed to be a function when $-1 < x < 1$? I realize there are an infinite number of values in that range, and an infinite number of them will be fine (and an infinite number of them won't), so that is a hand-wavey restriction I've set, but let me clarify with an example. If $f(x) = 4^x$, then $f(1/2) = \{-2, 2\}$. This clearly can't be within the domain a function, since it breaks the rule that every input gives a single output. What makes those values okay to be considered within the exponential function domain?
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1We always take the positive number, by convention/definition. So $4^{1/2}=2$ is unambiguous. – Adam Hughes Oct 05 '15 at 01:10
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Who said $4^{1/2} = -2$? The typical construction of the exponential function is as follows: Let $f: \mathbb{R} \to \mathbb{R}$ be a continuous function such that $f(0) = 1$ and for all $x,y \in \mathbb R$, $f(x+y) = f(x)f(y)$. One can easily show that $f(x) = f(1)^x$ for rational $x$ and use the fact the rationals are dense to extend it to the reals. – user217285 Oct 05 '15 at 01:10
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$a^{\frac{1}{2}}$ is not a set of numbers. Usually $a^b$ is defined to be $\exp(\ln_π(a)b)$ when $a,b \in \mathbb{C}$, where $\ln_π$ is the principal branch of the complex logarithm. In more basic mathematics $a^b$ can be defined as $a^b = \sup(\{ a^r : r \in \mathbb{Q} \})$ but only when $a > 0$ and $b \in \mathbb{R}$. Also, high-school mathematics might consider $(-a)^\frac{1}{n} = -(a^\frac{1}{n})$ for $a > 0$ and odd integer $n$, but that actually conflicts with the more general definition for complex numbers, so you cannot have both.
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