Section 1.1 Exponential Functions
References References
- Calculus, Early Transcendentals by Stewart, Section 1.4.
A function like \(f(x) = 2^x\) is called exponential because the argument \(x\) is the exponent in the formula. Exponential functions are the most basic and common transcendental functions, and are probably the most important functions in mathematics and science after polynomials.
We will see how exponential functions can be defined to have graphs that are continuous, unbroken curves with well defined slopes, rather being only a collection of separate points for integer values of \(x\text{.}\)
Natural number powers of 2.
With basic algebra, exponential functions are defined first for positive integer arguments, by formula
\begin{equation*}
2^n = 2 \cdot 2 \cdot 2 \cdots 2 \text{, the product of }n\text{ copies of }2.
\end{equation*}
Then to satisfy the rule \(2^{n+m} = 2^n \cdot 2^m\) for the case \(m = 0\) requires
\begin{equation*}
2^0 = 1
\end{equation*}
so all non-negative integers \(n\) are covered.Negative integer powers of 2.
For a negative integer \(n\text{,}\) \(|n|=-n\) is positive, and to satisfy the rule \(2^n \cdot 2^m = 2^{n+m}\) we must have \(2^n.2^{-n}=2^{n+(-n)}=2^0=1\text{,}\) and so dividing by \(2^{-n}=2^{|n|}\text{,}\)
\begin{equation*}
2^n=\frac{1}{2^{-n}}=\frac{1}{2^{|n|}} \quad \text{ for } n \text{ a negative integer.}
\end{equation*}
Rational powers of 2.
Next we can make sense of exponentials for rational exponents. To get the exponential \(2^r\) for any rational number \(r\) start with exponent \(1/q\text{,}\) \(q\) a positive integer. To satisfy the rule \((2^a)^b=2^{ab}\) requires \((2^{1/q})^q=2^{q/q}=2^{1}=2\text{,}\) so taking the \(q\)-th root of both sides of this equation,
\begin{equation*}
2^{1/q}=\sqrt[q]{2} \; \text{(the q-th root of 2) for q a positive integer.}
\end{equation*}
Finally, any rational number can be written as \(r=p/q\) with \(p\) an integer, \(q\) a positive integer, and the same rule requires
\begin{equation*}
2^{p/q}=((2^p)^{1/q})=\sqrt[q]{2^p}.
\end{equation*}
Irrational powers of 2 (so all power of 2).
The graph of \(2^x\) for all rational \(x\) looks like a dense collection of dots along a curve which increases to the right. Can we fill in the gaps at irrational values of \(x\) and get a smooth, uninterrupted curve? For example, can we make sense of an irrational power like \(2^{\sqrt{3}}\text{?}\)
A number like \(\sqrt{3}=1.73205\dots\) is approximated by a succession of decimal fractions \(1\text{,}\) \(1.7\text{,}\) \(1.73\text{,}\) \(1.732\text{,}\) \(1.7320\text{,}\) \(1.73205\) and so on: it is the limit of this sequence of rational numbers.
Raising 2 to each of these powers gives the following new sequence of numbers (everything rounded to five decimal places):
\begin{equation*}
2^1=2 \lt 2^{1.7}=3.24900 \lt 2^{1.73}=3.31727 \lt 2^{1.732}= 3.32188 \lt 2^{1.73205}=3.32200 \dots
\end{equation*}
All of these should be less that \(2^{\sqrt{3}}\) since the values are increasing as the exponent increases and \(\sqrt{3}\) is greater than each of these exponents.
On the other hand if we round up the decimal approximations of \(\sqrt{3}\text{,}\) the exponentials should all be greater than \(2^{\sqrt{3}}\text{:}\)
\begin{equation*}
2^2=4 \gt 2^{1.8}= 3.48220 \gt 2^{1.74}= 3.34035 \gt 2^{1.733} = 3.32418 \gt 2^{1.7321}= 3.32211 \gt 2^{1.73206}= 3.32202 \dots
\end{equation*}
It appears that
\begin{equation*}
2^{1.73205}=3.32200 \lt 2^{\sqrt{3}} \lt 3.32202 = 2^{1.73206},
\end{equation*}
so that \(2^{\sqrt{3}}\) rounded to four decimal places is \(3.3220\text{.}\) We could continue with either sequence to compute a value for \(2^{\sqrt{3}}\) to as many decimal places as we wish.
In this way, we can make sense of, and compute, any power of 2, rational or irrational, so we have made sense of the exponential function \(f(x)=2^x\) for all real arguments \(x\text{.}\)
Any power of any positive number.
There is nothing special about the base 2 used above except that it is positive: we could do the same thing with any positive real number \(a\text{,}\) to compute the exponential function \(f(x)=a^x\text{.}\) The graphs for the different functions vary mostly in that they are- increasing for \(a>1\text{,}\) and increase faster for larger values of \(a\text{,}\)
- decreasing for \(0 \lt a \lt 1\) and decrease faster for smaller values of \(a\text{,}\)
- and in the borderline case of \(a=1\text{,}\) the graph is a constant: \(1^x=1\text{.}\)
Rules for exponential functions.
The familiar rules for exponentials still hold just as with with rational exponents: for \(a\) and \(b\) positive and any real numbers \(x\) and \(y\text{,}\)- \(a^{x+y} = a^x \cdot a^y\text{,}\) and \(a^{x-y} = a^x / a^y\)
- \(\displaystyle (a^x)^y =a^{x \cdot y}\)
- \(\displaystyle (a \cdot b)^x =a^x \cdot b^x\)
Applications of exponential functions.
See Example 1.33 on Bacterial Growth in Section 1.5 2 of our text, OpenStax Calculus Volume 1.
Example 1.1.2. Radioactive Decay.
The half-life of strontium-90, \({}^{90}\)Sr, is 25 years. This means that half of any given quantity of \({}^{90}\)Sr will disintegrate in 25 years.
- If a sample of \({}^{90}\)Sr initially has a mass of 24mg, find an expression for the mass \(m(t)\) that remains after \(t\) years.
- Find the mass remaining after 40 years, correct to the nearest milligram.
- Use a graphing device to graph \(m(t)\) and use the graph to estimate the time required for the mass to be reduced to 5 mg.
The number \(e\).
Of all possible choice of the base \(a\) of an exponential function \(a^{x}\text{,}\) one is most convenient for mathematics because it makes the slope of the graph simplest: the number called \(e\text{,}\) with value approximately \(e \approx 2.71828\text{.}\)
The graphs of all exponential functions pass through the point \(P(0,1)\) on the \(y\)-axis, but the bigger \(a\) is, the faster the function value grows as \(x\) increases, so the greater the slope is at this point. The slope is zero for \(a=1\text{,}\) when the function is constant, and increases as \(a\) increases.
Experimenting with a graphing calculator suggests that the slope is less than 1 for \(2^{x}\text{,}\) but greater than 1 for \(3^{x}\text{.}\) So it seems that by increasing \(a\) to somewhere between 2 and 3, the slope will be 1 at \(P(0,1)\text{,}\) with the slope greater than 1 for greater values of \(a\text{,}\) less than 1 for lesser values. That is, there is one special value for the base that gives slope 1: this is the value called \(e\text{.}\)
We have already seen that \(e\) lies between 2 and 3, and with ever more careful computation of slopes we could calculate the more accurate value given above.
We will soon see that any other exponential function can be written in terms of \(e^x\text{,}\) and this is very convenient in calculus, making this particular exponential function so important that it is often called simply “the exponential function”.
There is another way to see the origins of this special number in terms of continuous growth, discussed in Section 1.5 3 of the text using the example of continuously compounded interest.
Exercises Exercises
Study Calculus Volume 1, Section 1.5 4 , Exercises 233, 235, 239, 243, 278, 279, 299, 301, and 305(a) [we will get to part (b) soon].
openstax.org/books/calculus-volume-1/pages/1-5-exponential-and-logarithmic-functionsopenstax.org/books/calculus-volume-1/pages/1-5-exponential-and-logarithmic-functionsopenstax.org/books/calculus-volume-1/pages/1-5-exponential-and-logarithmic-functionsopenstax.org/books/calculus-volume-1/pages/1-5-exponential-and-logarithmic-functions
