That's what the Lord told Adam and Eve on the sixth day of creation according to Genesis 1:28. Again in Genesis 22:17, the Lord tells Abraham, "I will multiply thy seed as the stars of the heaven, and as the sand which is upon the seashore."

Clearly as far back as Abraham, ancient peoples had some understanding of exponential functions. And Abraham lived ever so long ago.

There is also this legend -- not quite as old as Genesis -- that by some brave deed a peasant saved all of Persia from its enemies. So pleased was the shah that he offered this peasant a piece of the kingdom. The peasant replied by saying, "Sire, my wants are modest. Simply give me a single grain of wheat for the first square of the chess board, two grains for the second square, four grains for the third, and so on, doubling the number of grains for each square until you reach the sixty-fourth square."

The shah agreed to this in front of witnesses and ordered the court accountant to compute how many sacks of grain that would be. Much to the shah's horror, there weren't that many sacks in all of Persia, much less that much grain. It turns out that if you allow 50 milligrams per grain of wheat, it would take 460 billion tons of wheat (in the ballpark of United States production over a thousand years) to meet the shah's obligation to the peasant.

For his ignorance of exponentials, the shah was embarassed before his subjects. Whether the peasant was ignorant or not of exponentials, we shall never know. He most certainly paid with his head for his impertinence.

One final example of an exponential. A certain species of bacteria has a generation time of 60 minutes. That means that a bacterium of this species allowed to exist under favorable conditions will, after 60 minutes, divide into two. 60 minutes later those two will divide into four. Another 60 minutes and those four will divide into eight, and so on.

But here is an extra wrinkle to the example. The generation time is not
always *exactly* 60 minutes for every bacterium. Sometimes it can be as short as
50 minutes or as long as 70 minutes. It only *averages* 60 minutes.
So even if you start out with the original bacterium dividing on the hour,
soon some of its descendants will be dividing not just on the hour, but
at all times throughout the hour.

We would still expect that after, say, 10 hours, we would find
very nearly 1024 (that is `2 ^{10}`) bacteria cells, and
at 11 hours we would expect to find very nearly 2048 (that is

We know from the fix the shah found himself in that within a few days the number of bacteria cells will grow very large indeed. And however many there are at any hour, there will be very nearly that many cell divisions over the subsequent hour. So within a few days there will be so many cell divisions each second that you could almost regard the cell divisions as being a continuous process. Any interval of time during which no cell division took place would be so short as to be imperceptible even by the most sensitive of time-measuring equipment.

And that is where we're going with the concept of exponentials.
We want to turn them into continuous functions.
We begin with the idea of raising a number to some
`n`th
power, which we do by multiplying it by itself `n` times.
Whenever `n` is a counting number, we can always do this.
But what about all those real numbers between `n` and `n+1`?
How do we raise a number to those powers? How do we make exponentials
into something continuous that we can do calculus on?

We now begin a review of what you learned in algebra about exponents.

Let's begin with the most elementary property of raising some number,
`b`, to a power. If `m` and `n` are counting numbers,
then we have by definition,

bFrom that it is quite clear that^{n}= b × b × ... n times ... × b eq. 6.1-1a b^{m}= b × b × ... m times ... × b eq. 6.1-1b

bWe infer from this the general rule (which you already know) that^{m+n}= b × b × ... m+n times ... × b = (b × b × ... m times ... × b) × (b × b × ... n times ... × b) eq. 6.1-2

bIn words this means that to take the product of some number raised to two powers, find the sum of the exponents and raise it to that power.^{m+n}= b^{m}× b^{n}eq. 6.1-3

This rule forms the foundation of our understanding of exponentials. As we develop exponentials into continuous functions, we shall keep falling back on this rule to guide our way. That is, we will want our extended definitions of exponentials to obey the rule just as our original definition did when we restricted exponents to being counting numbers.

For example, if we want to discover what `b ^{0}` is, then
we simply look at what happens to the rule when we take

bDividing both the left expression and the right expression by^{n}= b^{n+0}= b^{n}× b^{0}eq. 6.1-3a

bThis works for any^{n}b^{n+0}1 =~~=~~~~= b~~^{0}eq. 6.1-3b b^{n}b^{n}

Likewise the rule leads us to a method on raising numbers to a negative
power. If `n` is a counting number, then by the rule we have:

1 = bDividing through by^{0}= b^{n+(-n)}= b^{n}× b^{-n}eq. 6.1-4a

1 bThat takes care of raising any nonzero^{0}b^{n+(-n)}~~=~~~~=~~~~= b~~^{-n}eq. 6.1-4b b^{n}b^{n}b^{n}

To deal with them we observe another consequence of the rule. If `p`
and `q` are counting numbers then we have from basic arithmetic:

pq = p + p + ... q times ... + p eq. 6.1-5So it must also be true that

bIf you apply the rule to the right-hand expression in this, you get:^{pq}= b^{(p + p + ... q times ... + p)}eq. 6.1-6a

bAnd by the original definition of how to raise a number to a power that is a counting number, the right-hand side of the above is the same as^{pq}= b^{p}× b^{p}× ... q times ... × b^{p}eq. 6.1-6b

bThis rule is a consequence of the first. And we shall extend it too into the realm of exponents that are not counting numbers.^{pq}= (b^{p})^{q}eq. 6.1-6c

Start out with the obvious:

bLet^{1}= b eq. 6.1-7

(bThis equation doesn't tell us what^{1/n})^{n}= b^{n/n}= b^{1}= b eq. 6.1-8

f(x) = xIn your algebra studies, they told you that whenever^{n}eq. 6.1-9

Does what we've learned about calculus so far provide any clues
about the behavior of the function,
`f(x) = x ^{n}`

Suppose that ` b > 0`` f(x) = x ^{n}`

f(x_{high}) > b eq. 6.1-10

That means we can bracket `f(x)` both above and below `b`.
That is, `0 < b < f(x _{high})`

f(x) = xAnd that solution lies somewhere on^{n}= b eq. 6.1-11

Not only that, but `f(x) = x ^{n}`

f'(x) = nxwhich is also continuous and always positive whenever^{n-1}eq. 6.1-12

And that means that *only a single* positive `x` can satisfy
`x ^{n} = b`

**end of optional material**

Clearly if for ` b > 0 `

bthen you satisfy equation 6.1-8. And if you restrict applying exponents only to positive numbers, then only the^{1/n}= nth root of b eq. 6.1-13

Clearly if `b ^{1/n}` is a positive real number whenever

(bYou can also readily see, by grouping terms, that^{1/n})^{m}= b^{m/n}eq. 6.1-14a

(band in general^{1/n})^{m}= (b^{1/2n})^{2m}eq. 6.1-14b

bfor any counting number,^{m/n}= b^{pm/pn}eq. 6.1-14c

m pmYou can make~~=~~~~n pn~~

That still leaves one problem. We know there are plenty of real numbers that aren't rationals. How do you take a positive real number to a real power when that power isn't a rational number?

There is only one answer, and that is to use limits.

Recall that our original goal was to have exponential functions be
continuous. Recall also that every real number, `r` is the limit of a
Cauchy sequence of rationals. That means that for some sequence of
rational numbers,
` q _{1}, q_{2}, q_{3}, ... `,

r = lim qIf_{k}eq. 6.1-15 k~~> ¥~~

bOn the right hand side of the equation, we have^{r}= lim b^{qk}eq. 6.1-16 k~~> ¥~~

Isaac Newton observed that a heated object left out to cool does so according to the following formula:

T(t) - Twhere_{a}= A b^{-t}

Let `t` be in minutes and `T(t)` be in degrees C. A loaf of bread
is pulled from the oven and allowed to cool. It's initial temperature when it
is pulled from the oven is 100 C. After 5 minutes it has cooled to 75 C. After
10 minutes it has cooled to 60 C. Find the ambient temperature of the room,
`T _{a}`, and the two constants,

**Step 1: What does the problem tell you about** `T(t)`**?**
It gives you three different temperatures at three different times. That is,
it tells you that at `t = 0`

T(0) = 100Likewise, it tells you that at

T(5) = 75 T(10) = 60respectively.

**Step 2: What equations can you make from this information?**
You can plug each of the values above into the original equation.
For ` t = 0 `

100 - TLikewise with the other two you can form the equations:_{a}= A b^{0}

75 - Tand_{a}= A b^{-5}

60 - T_{a}= A b^{-10}

**Step 3: How can you eliminate a variable from these equations?**
Start out with the easiest equation, which is the first one.
You know that no matter
what `b` is, ` b ^{0} = 1`

100 - TThat means that in the other two equations, everywhere you see_{a}= A

75 - T_{a}= (100 - T_{a})b^{-5}60 - T_{a}= (100 - T_{a})b^{-10}

**Step 4: How can you eliminate yet another variable from the above two
equations?**
There are several approaches at this point, but I will show you what I think
is the easiest. If you divide both equations through by
` (100 - T _{a})`

75 - TBut observe that_{a}~~= b~~^{-5}100 - T_{a}60 - T_{a}~~= b~~^{-10}100 - T_{a}

60 - Tor equivalently_{a}(75 - T_{a})^{2}~~=~~~~100 - T~~_{a}(100 - T_{a})^{2}

(60 - T_{a})(100 - T_{a}) = (75 - T_{a})^{2}

**Step 5: Multiply it out and solve for** `T _{a}`.

6000 - 160TThe_{a}+ T_{a}^{2}= 5625 - 150T_{a}+ T_{a}^{2}

375 = 10T_{a}37.5 = T_{a}

**Step 6: Back Substitute to find **`b ^{-5}`.
You now know what

75 - 37.5 37.5 3So~~=~~~~=~~~~= b~~^{-5}100 - 37.5 62.5 5

I leave it up to you to plug these numbers back into the original problem to see that they do indeed work.

How much have you learned about the behavior of exponentials? Using methods similar to those in the text, can you prove that

awhen^{r}× b^{r}= (ab)^{r}

Start out by showing it is true for

awhen^{n}× b^{n}= (ab)^{n}

Next use a little algebra to show that it is true for

aas well by using the definition given in eq. 6.1-4b.^{-n}× b^{-n}= (ab)^{-n}

From there demonstrate that

aThis is a little more difficult, but see where you can get to starting with^{1/n}× b^{1/n}= (ab)^{1/n}

(awhich is a consequence of the first part of this exercise. Then apply what we know about the product of exponents.^{1/n})^{n}× (b^{1/n})^{n}= (a^{1/n}b^{1/n})^{n}

Once you have that, it should be easy to show that

awhenever^{m/n}× b^{m/n}= (ab)^{m/n}

Finally, use the continuity of exponentials (and the product rule for limits) to argue that

afor any real number,^{r}× b^{r}= (ab)^{r}

Move on to The Farther We Go, The Faster We Get There (derivative of exponentials)

email me at *hahn@netsrq.com*