This chapter is devoted to exponentials like 2x and 10x and above all ex. The goal is to understand them, differentiate them, integrate them, solve equations with them, and invert them (to reach the logarithm). The overwhelming importance of ex makes this a crucial chapter in pure and applied mathematics.
In the traditional order of calculus books, ex waits until other applications of the integral are complete. I would like to explain why it is placed earlier here. I believe that the equation dy/dx = y has to be emphasized above techniques of integration. The laws of nature are expressed by differential equations, and at the center is ex. Its applications are to life sciences and physical sciences and economics and engineering (and more—wherever change is influenced by the present state). The model produces a differential equation and I want to show what calculus can do.
The key is always bm+n = (bm)(bn). Section 6.1 applies that rule in three ways:
There is a good chance you have met logarithms. They turn multiplication into addition, which is a lot simpler. They are the basis for slide rules (not so important) and for graphs on log paper (very important). Logarithms are mirror images of exponentials—and those I know you have met.
Start with exponentials. The numbers 10 and 10² and 10³ are basic to the decimal system. For completeness I also include 10⁰, which is "ten to the zeroth power" or 1. The logarithms of those numbers are the exponents. The logarithms of 1 and 10 and 100 and 1000 are 0 and 1 and 2 and 3. These are logarithms "to base 10,"because the powers are powers of 10.
Question When the base changes from 10 to b, what is the logarithm of 1?
Answer Since b0 = 1, logb1 is always zero. To base b, the logarithm of bn is n.
Negative powers are also needed. The number 10x is positive, but its exponent x can be negative. The first examples are 1/10 and 1/100, which are the same as 10-1 and 10-2. The logarithms are the exponents -1 and -2:
1000 = 10³ and log 1000 = 3
1/1000 = 10-3 and log 1/1000 = - 3.
Multiplying 1000 times 1/1000 gives 1 = 100. Adding logarithms gives 3 + (−3) = 0. Always 10m times 10n equals 10m+n. In particular 10³ times 10² produces five tens:
(10)(10)(10) times (10)(10) equals (10)(10)(10)(10)(10) = 105.
The law for bm times bn extends to all exponents, as in 104.6 times 10π. Furthermore the law applies to all bases (we restrict the base to b > 0 and b ≠ 1). In every case multiplication of numbers is addition of exponents.
6A bm times bn equals bm+n, so logarithms (exponents) add bm divided by bn equals bm-n, so logarithms (exponents) subtract
logb(yz) = logby + 10gbz and logb(y/z) = logby − logbz. (1)
Historical note In the days of slide rules, 1.2 and 1.3 were multiplied by sliding one edge across to 1.2 and reading the answer under 1.3. A slide rule made in Germany would give the third digit in 1.56. Its photograph shows the numbers on a log scale. The distance from 1 to 2 equals the distance from 2 to 4 and from 4 to 8. By sliding the edges, you add distances and multiply numbers.
Division goes the other way. Notice how 1000/10 = 100 matches 3 - 1 = 2. To divide 1.56 by 1.3, look back along line D for the answer 1.2.
The second figure, though smaller, is the important one. When x increases by 1, 2x is multiplied by 2. Adding to x multiplies y. This rule easily gives y = 1, 2, 4, 8, but look ahead to calculus—which doesn't stay with whole numbers.
Calculus will add Δx. Then y is multiplied by 2ax. This number is near 1. If Δx = 1/10 then 2Δx ≈ 1.07—the tenth root of 2. To find the slope, we have to consider (2Δx − 1)/Δx. The limit is near (1.07 − 1)/(1/10) = .7, but the exact number will take time.
Base Change Bases other than 10 and exponents other than 1,2,3, … are needed for applications. The population of the world x years from now is predicted to grow by a factor close to 1.02x. Certainly x does not need to be a whole number of years. And certainly the base 1.02 should not be 10 (or we are in real trouble). This prediction will be refined as we study the differential equations for growth. It can be rewritten to base 10 if that is preferred (but look at the exponent):
1.02x is the same as 10(log 1.02)x.
When the base changes from 1.02 to 10, the exponent is multiplied—as we now see.
For practice, start with base b and change to base a. The logarithm to base a will be written "log." Everything comes from the rule that logarithm = exponent:
base change for numbers: b = alog b
Now raise both sides to the power x. You see the change in the exponent:
base change for exponentials: bx = a(log b)x
Finally set y = bx. Its logarithm to base b is x. Its logarithm to base a is the exponent on the right hand side: loga y = (loga b)x. Now replace x by logb y:
base change for logarithms: loga y = (loga b)(logb y ).
We absolutely need this ability to change the base. An example with a = 2 is
b = 8 = 2³ 8² = (2³)² = 26 log2 64 = 3⋅2 = (log2 8)(log8 64).
The rule behind base changes is (am)x = amx. When the mth power is raised to the xth power, the exponents multiply. The square of the cube is the sixth power:
(a)(a)(a) times (a)(a)(a) equals (a)(a)(a)(a)(a)(a): (a³)² = a6.
Another base will soon be more important than 10—here are the rules for base changes:
6B To change numbers, powers, and logarithms from base b to base a, use
b = aloga b bx = a(loga b)x loga y = (loga b)(logb y) (2)
The first is the definition. The second is the xth power of the first. The third is the logarithm of the second (remember y is bx). An important case is y = a:
loga a = (loga b)(logb a) = 1 so loga b = 1/logb a. (3)
This completes the algebra of logarithms. The addition rules 6A came from (bm)(bn) = bm+n. The multiplication rule 6B came from (am)x = amx. We still need to deJine bx and ax for all real numbers x. When x is a fraction, the definition is easy. The square root of a8 is a4 (m = 8 times x = 1/2). When x is not a fraction, as in 2π, the graph suggests one way to fill in the hole.
We could define 2π as the limit of 23, 231/10, 2314/100, …. As the fractions r approach π, the powers 2r approach 2π. This makes y = 2x into a continuous function, with the desired properties (2m)(2n) = 2m+n and (2m)x = 2mx —whether m and n and x are integers or not. But the ε's and δ's of continuity are not attractive, and we eventually choose (in Section 6.4) a smoother approach based on integrals.
It is time to draw graphs. In principle one graph should do the job for both functions, because y = bx means the same as x = logb y. These are inverse functions. What one function does, its inverse undoes. The logarithm of g(x) = bx is x:
g-1(g(x)) = logb (bx) = x. (4)
In the opposite direction, the exponential of the logarithm of y is y:
g(g(-1y)) = b(logb y) = y. (5)
This holds for every base b, and it is valuable to see b = 2 and b = 4 on the same graph. Figure 6.2a shows y = 2x and y = 4x. Their mirror images in the 45° line give the logarithms to base 2 and base 4, which are in the right graph.
When x is negative, y = bx is still positive. If the first graph is extended to the left, it stays above the x axis. Sketch it in with your pencil. Also extend the second graph down, to be the mirror image. Don't cross the vertical axis.
There are interesting relations within the left figure. All exponentials start at 1, because b0 is always 1. At the height y = 16, one graph is above x = 2 (because 4² = 16). The other graph is above x = 4 (because 24 = 16). Why does 4x in one graph equal 22x in the other? This is the base change for powers, since 4 = 2².
The figure on the right shows the mirror image—the logarithm. All logarithms start from zero at y = 1. The graphs go down to −∞ at y = 0. (Roughly speaking 2−∞ is zero.) Again x in one graph corresponds to 2x in the other (base change for logarithms). Both logarithms climb slowly, since the exponentials climb so fast.
The number log2 10 is between 3 and 4, because 10 is between 2³ and 24. The slope of 2x is proportional to 2x —which never happened for xn. But there are two practical difficulties with those graphs:
There is also another point. In many problems we don't know the function y = ƒ(x). We are looking for it! All we have are measured values of y (with errors mixed in). When the values are plotted on a graph, we want to discover ƒ(x).
Fortunately there is a solution. Scale the y axis differently. On ordinary graphs, each unit upward adds a fixed amount to y. On a log scale each unit multiplies y by a fixed amount. The step from y = 1 to y = 2 is the same length as the step from 3 to 6 or 10 to 20.
On a log scale, y = 11 is not halfway between 10 and 12. And y = 0 is not there at all. Each step down divides by a fixed amount—we never reach zero. This is completely satisfactory for Abx, which also never reaches zero.
Figure 6.3 is on semilog paper (also known as log-linear), with an ordinary x axis. The graph of y = Abx is a straight line. To see why, take logarithms of that equation:
log y = log A + x log b. (6)
The relation between x and log y is linear. It is really log y that is plotted, so the graph is straight. The markings on the y axis allow you to enter y without looking up its logarithm—you get an ordinary graph of log y against x.
Figure 6.3 shows two examples. One graph is an exact plot of y = 2⋅10x. It goes upward with slope 1, because a unit across has the same length as multiplication by 10 going up. 10x has slope 1 and 10(log b)x (which is bx) will have slope log b. The crucial number log b can be measured directly as the slope.
The second graph in Figure 6.3 is more typical of actual practice, in which we start with measurements and look for ƒ(x). Here are the data points:
We don't know in advance whether these values fit the model y = Abx. The graph is strong evidence that they do. The points lie close to a line with negative slope—indicating log b < 0 and b < 1. The slope down is half of the earlier slope up, so the model is consistent with
y = A⋅10-x/2 or log y=log A − ½x. (7)
When x reaches 2, y drops by a factor of 10. At x = 0 we see A ≈ 4.
Another model—a power y = Axk instead of an exponential—also stands out with logarithmic scaling. This time we use log-log paper, with both axes scaled. The logarithm of y = Axk gives a linear relation between log y and log x:
log y = log A + k log x. (8)
The graphs in Figure 6.4 have slopes 3 and ½ and -1. They represent Ax³ and A√x and A/x. To find the A's, look at one point on the line. At x = 4 the height is 8, so adjust the A's to make this happen: The functions are x³/8 and 4√x and 32/x. On semilog paper those graphs would not be straight!
You can buy log paper or create it with computer graphics.
This is a calculus book. We have to ask about slopes. The algebra of exponents is done, the rules are set, and on log paper the graphs are straight. Now come limits.
The central question is the derivative. What is dy/dx when y = bx? What is dx/dy when x is the logarithm logb y? Thpse questions are closely related, because bx and logby are inverse functions. If one slope can be found, the other is known from dx/dy = 1/(dy/dx). The problem is to find one of them, and the exponential comes first.
You will now see that those questions have quick (and beautiful) answers, except for a mysterious constant. There is a multiplying factor c which needs more time. I think it is worth separating out the part that can be done immediately, leaving c in dyldx and llc in dx/dy. Then Section 6.2 discovers c by studying the special number called e (but c ≠ e).
6C The derivative of bx is a multiple cbx. The number c depends on the base b.
The product and power and chain rules do not yield this derivative. We are pushed all the way back to the original definition, the limit of Δy/Δx:
dy/dx = limh→0 (y(x + h) − y(x)) / h = limh→0 (bx+h − bx) / h. (9)
Key idea: Split bx+h into bx times bh. Then the crucial quantity bx factors out. More than that, bx comes outside the limit because it does not depend on h. The remaining limit, inside the brackets, is the number c that we don't yet know:
| dy | = limh→0 | bxbh − bx | = bx | limh→0 | bk−1 | = cbx. (10) | ||
| dx | h | h |
This equation is central to the whole chapter: dy/dx equals cbx which equals cy. The rate of change of y is proportional to y. The slope increases in the same way that bx increases (except for the factor c). A typical example is money in a bank, where interest is proportional to the principal. The rich get richer, and the poor get slightly richer. We will come back to compound interest, and identify b and c.
The inverse function is x = logb y. Now the unknown factor is 1/c:
6D The slope of logb y is 1/cy with the same e (depending on b).
Proof If dy/dx = cbx then dx/dy = 1/cbx = 1/cy. (11)
That proof was like a Russian toast, powerful but too quick! We go more carefully:
The logarithm gives another way to find c. From its slope we can discover 1/c. This is the way that finally works (next section).
Final remark It is extremely satisfying to meet an ƒ(y) whose derivative is 1/cy. At last the "−1 power" has an antiderivative. Remember that ∫ xn dx = xn+1/(n + 1) is a failure when n = −1. The derivative of x0 (a constant) does not produce x-1. We had no integral for x-1, and the logarithm fills that gap. If y is replaced by x or t (all dummy variables) then
| d | logb x = | 1 | and | d | logb t = | 1 | . (12) |
| dx | cx | dt | ct |
The base b can be chosen so that c = 1. Then the derivative is 1/x. This final touch comes from the magic choice b = e —the highlight of Section 6.2.
