Calculus Early Transcendentals: Differential & Multi-Variable Calculus for Social Sciences

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 “yes.”

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 “obvious” 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 “holes” 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 “filled in”, 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 \(0\lt a\lt 1\text{:}\)