丁香园AV

An exponential function is a function of the form \(f(x)=a^x\text{,}\) where \(a\) is a constant. Examples are \(2^x\text{,}\) \(10^x\) and \((1/2)^x\text{.}\) To more formally define the exponential function we look at various kinds of input values.

It is obvious that \(\ds a^5=a\cdot a\cdot a\cdot a\cdot a\) and \(\ds a^3=a\cdot a\cdot a\text{,}\) but when we consider an exponential function \(\ds a^x\) we can't be limited to substituting integers for \(x\text{.}\) What does \(\ds a^{2.5}\) or \(a^{-1.3}\) or \(\ds a^\pi\) mean? And is it really true that \(a^{2.5}a^{-1.3}=a^{2.5-1.3}\text{?}\) The answer to the first question is actually quite difficult, so we will evade it; the answer to the second question is 鈥測es.鈥�

We'll evade the full answer to the hard question, but we have to know something about exponential functions. You need first to understand that since it's not 鈥渙bvious鈥� what \(\ds 2^x\) should mean, we are really free to make it mean whatever we want, so long as we keep the behavior that is obvious, namely, when \(x\) is a positive integer. What else do we want to be true about \(\ds 2^x\text{?}\) We want the properties of the previous two paragraphs to be true for all exponents: \(\ds 2^x2^y=2^{x+y}\) and \(\ds (2^x)^y=2^{xy}\text{.}\)

After the positive integers, the next easiest number to understand is 0: \(\ds 2^0=1\text{.}\) You have presumably learned this fact in the past; why is it true? It is true precisely because we want \(\ds 2^a2^b=2^{a+b}\) to be true about the function \(\ds 2^x\text{.}\) We need it to be true that \(\ds 2^02^x=2^{0+x}=2^x\text{,}\) and this only works if \(\ds 2^0=1\text{.}\) The same argument implies that \(\ds a^0=1\) for any \(a\text{.}\)

The next easiest set of numbers to understand is the negative integers: for example, \(\ds 2^{-3}=1/2^3\text{.}\) We know that whatever \(\ds 2^{-3}\) means it must be that \(\ds 2^{-3}2^{3}=2^{-3+3}=2^0=1\text{,}\) which means that \(\ds 2^{-3}\) must be \(1/2^3\text{.}\) In fact, by the same argument, once we know what \(\ds 2^x\) means for some value of \(x\text{,}\) \(\ds 2^{-x}\) must be \(\ds 1/2^{x}\) and more generally \(a^{-x}=1/a^x\text{.}\)

Next, consider an exponent \(1/q\text{,}\) where \(q\) is a positive integer. We want it to be true that \(\ds (2^x)^y=2^{xy}\text{,}\) so \(\ds (2^{1/q})^q=2\text{.}\) This means that \(\ds 2^{1/q}\) is a \(q\)-th root of 2, \(\ds 2^{1/q}=\root q\of{2\ }\text{.}\) This is all we need to understand that \(2^{p/q}=(2^{1/q})^p=(\root q\of{2\ })^p\) and \(a^{p/q}=(a^{1/q})^p=(\root q\of{a\ })^p\text{.}\)

What's left is the hard part: what does \(\ds 2^x\) mean when \(x\) cannot be written as a fraction, like \(\ds x=\sqrt{2\ }\) or \(\ds x=\pi\text{?}\) What we know so far is how to assign meaning to \(\ds 2^x\) whenever \(x=p/q\text{.}\) If we were to graph \(a^x\) (for some \(a>1\)) at points \(x=p/q\) then we'd see something like this:

This is a poor picture, but it illustrates a series of individual points above the rational numbers on the \(x\)-axis. There are really a lot of 鈥渉oles鈥� in the curve, above \(x=\pi\text{,}\) for example. But (this is the hard part) it is possible to prove that the holes can be 鈥渇illed in鈥�, and that the resulting function, called \(\ds a^x\text{,}\) really does have the properties we want, namely that \(\ds a^xa^y=a^{x+y}\) and \(\ds (a^x)^y=a^{xy}\text{.}\) Such a graph would then look like this:

Three Types of Exponential Functions.

There are three kinds of exponential functions \(f(x)=a^x\) with \(x\) real depending on whether \(a>1\text{,}\) \(a=1\) or \(0\lt a\lt 1\text{:}\)