# Section 6: Exponentials and Logarithms

© 1997 by Karl Hahn

## 6.3 Be Frugal and Add (logarithms)

In all likelihood, you have already studied logarithms when you took algebra. At that time, they probably seemed useful only for carrying out difficult computations back in the days before electronic computers. But in the ediface of calculus, logarithms in general, and the natural log function in particular, are a keystone in our understanding. Because they are so essential to calculus, they turn up everywhere in physics, chemistry, and engineering. Even with modern computers to help us calculate, we still wouldn't get very far without logarithms.

In the last section we defined the natural log function, ln(x), as

```                      xh - 1
ln(x)  =    lim                                                eq. 6.3-1
h  > 0     h
```
where x > 0.

That definition of the natural log function arose because of the role it plays in finding the derivative of  f(x) = bx. In the problem set for the last section, you used that role along with the familiar properties of exponentials to find that the ln function obeys the rules of logarithms. That is:

```   ln(ab)  =  ln(a) + ln(b)                                       eq. 6.3-2a
```
and
```   ln(bn)  =  n ln(b)                                             eq. 6.3-2b
```
And you would expect it to obey those rules -- otherwise naming it the natural log would have been inappropriate, wouldn't it? These very properties, though, can be deduced from the limit definition alone. Those derivations are given here as a
coached exercise, and although you are unlikely to encounter these on an exam, you should follow and understand them, since it will improve your strength at doing limits problems.

### Review of Logs

When they introduced you to logarithms in algebra, they probably talked about the base of a logarithm. Usually you study logs base 10 before any others. And the symbol for the log base 10 of x was log10(x). Recall now that for log10 you had

```   .
.
.
log10(0.0001)  = -4
log10(0.001)   = -3
log10(0.01)    = -2
log10(0.1)     = -1
log10(1)       =  0
log10(10)      =  1
log10(100)     =  2
log10(1000)    =  3
log10(1000)    =  4
.
.
.

table 6.3-1
```

Figure 6.3-1 shows a plot of  f(x) = 10 log10(x). I have multiplied it by 10 to stretch in the vertical direction so that you can more easily view its behavior between  x = 1  and  x = 10.

Logs base 10 obey the same product rule as natural logs. That is log10(ab) = log10(a) + log10(b). So if you needed to know the log base 10 of 40000, you could look on the graph and know that  log10(4) = 0.6021  and look at table 6.3-1 and know that  log10(10000) = 4, and by adding them, you would know that  log10(40000) = 4.6021. And by the way, the logs base 10 given in the graph for 2 through 9 are only approximate to four decimal places. All of those values are, in truth, irrational. The values for the logs base 10 of 1 and 10, though, are exactly 0 and 1 respectively.

You probably also learned that  f(x) = 10x  and  g(x) = log10(x)  are inverse functions of each other. That means that

```   log10( 10x )  =  x
```
and
```   10log10(x)  =  x
```
Now that you know that exponentials are always continuous functions, and that they are monotone (that is either always increasing or always decreasing), You can use The Intermediate Value Theorem to show that the inverse always exists. For if you were given a positive x and asked to look for the y that solves
```   x  =  10y
```
you could find a yhigh such that
```   x  <  10yhigh
```
by choosing yhigh big enough and a ylow such that
```   x  >  10ylow
```
by choosing ylow small enough. Since 10y is a continuous function of y, the intermediate value theorem demands the existence of a y that makes x = 10y true. And that means that if log10 is the inverse function of 10x, then y = log10(x) exists for any positive x anybody might name.

You can also readily show that log10(x) is monotone increasing. It follows from 10x being monotone increasing. If log10(x) were not monotone increasing, then their would be some x1 > x2 such that  log10(x1) £ log10(x2). But if you take 10log10(x1) and 10log10(x2), you get back x1 and x2 respectively, and 10x being monotone would require that  x1 £ x2. Again we have a proof by contradiction.

And what about logs using bases other than 10? Suppose you wanted log base 7 of x. Observe that

```   x  =  7log7(x)                                                 eq. 6.3-3a
```
which follows from log and exponentiation being inverses of each other. If you take the log base 10 of both sides and apply what we know about logs, you have
```   log10(x)  =  log7(x) × log10(7)                                eq. 6.3-3b
```
Divide out log10(7) and you have
```               log10(x)
log7(x)  =                                                     eq. 6.3-3c
log10(7)
```
And in general, if you have log to the base a of x, you can convert it to log to the base b of x by this formula:
```               loga(x)
logb(x)  =                                                     eq. 6.3-4
loga(b)
```
where a, b, and x can be any positive real numbers.

Here is one more property of logs that is important. If you take the log to the base b, where b is any positive real number, then

```   logb(b)  =  1                                                  eq. 6.3-5
```
You should be able to verify this yourself by raising b to the power on each side of equation 6.3-5 and remembering that logs and exponentials are inverses of each other.

### The Base of the Natural Log

This has all been review of material you had in algebra so far. Now let's get down to what logs in general and natural log in particular have to do with calculus. Remember that we defined the following limit as natural log:

```                    xh - 1
ln(x)  =   lim                                                 eq. 6.3-6
h  > 0    h
```
If the function, ln(x), is truly a log in the sense that log to the base 10 and log to the base 7 are logs, then ln(x) must have a base as well. That means that there must be a positive real number, e, such that for all positive x it is true that
```   ln(x)  =  loge(x)                                              eq. 6.3-7
```
And from equation 6.3-5, we also have
```                    eh - 1
ln(e)  =   lim           =  loge(e)  =  1                      eq. 6.3-8
h  > 0    h
```
It turns out that the real number that satisfies this equation is irrational. To 16 places,  e = 2.718281828459045. But is the limit we used to define ln(x) in eq. 6.3-6 indeed the inverse function of ex? Here's proof (I can't tell you if this proof will be on any exam. It depends upon your instructor): The two functions can be inverses of each other if and only if  ln(ex) = x  for all x.
```                     (ex)h - 1            ehx - 1
ln(ex)  =   lim              =   lim                           eq. 6.3-9a
h  > 0     h         h  > 0    h
```
Now we employ a trick we've used before. We substitute k for hx. If h goes to zero, then k certainly must go to zero also. After all, we have  k = hx. By the way, that also means you can substitute the h in the denominator with k/x. Why? Because  h = k/x. With those substitutions and just a little algebra, eq. 6.3-9a becomes
```                       ek - 1
ln(ex)  =   lim   x                                            eq. 6.3-9b
k  > 0      k
```
But we already know that
```          ek - 1
lim           =  1                                            eq. 6.3-9c
k  > 0    k
```
That is because we defined e to be the real number that solves this equation. So we are left with
```   ln(ex)  =   lim   x × 1  =  x                                  eq. 6.3-9d
k  > 0
```
which finishes the proof and confirms that ln(x) and ex are inverses of each other. It also shows that natural log of x is the same thing as log to the base e of x.

### A Very Special Property

Now if  ln(e) = 1, what do you suppose the derivative of  f(x) = ex  is? Simply apply the formula we developed in the last section for taking the derivative of an exponential.

```   f'(x)  =  ln(e) × ex  =  ex                                    eq. 6.3-10
```
That's right. The function f(x) = ex is its own derivative. This is the operative property of the ex function. For the purpose of doing calculus, this property makes ex the mother of all exponential functions. Indeed, in many problems that involve exponentials other than ex, the first step is to convert it to an ex problem using the following identity:
```   bx  =  ex ln(b)                                                eq. 6.3-11
```
which holds for any positive real number, b, and any real number, x. You should be able to verify this identity for yourself based upon the discussion so far. If you can't, you should go back and review until you can.

A few other things you should know about ex. Like other exponential functions, it is defined and continuous for all real x. And like other exponential functions, the value of ex is positive for all real x. Just to emphasize that last remark, if x is any real number, ex is never zero and never negative. Another useful fact is that  e0 = 1  in the same way a zero exponent causes any exponential to be 1.

Figure 6.3-2 shows a plot of ex. Observe that this function always returns a result that is positive. Like other exponential functions, it crosses the y-axis at  y = 1. Notice that it crosses  x = 1  at  y = 2.718281728. Notice that as x becomes more and more positive, ex gets very big very fast. And as x becomes more and more negative, ex gets very close to zero very fast. Also, if you eyeball the curve carefully, you can see that its slope as it crosses the y-axis is 1.

### Derivative of the Natural Log

The function,  f(x) = ln(x), has a derivative as well, and it is important that you know it and know why its derivative is what it is. I will cover two different derivations of this derivative. If you are asked to demonstrate the derivative of the natural log on an exam, pay attention to the wording of the problem. It may contain a clue as to which of these derivations the instructor wants to see.

We'll do the easy one first. We know from the discussion so far that ln(x) and ex are inverses of each other. That means that if f(x) = ln(x), then

```   ef(x)  =  eln(x)  =  x                                         eq. 6.3-12
```
Now remember that ex is its own derivative. Using the chain rule to take the derivative of the left hand term (we'll ignore the middle term), and elementary derivatives to take the derivative of the right hand term, we have
```   ef(x) f'(x)  =  1                                              eq. 6.3-13a
```
But from eq. 6.3-12, we know that  ef(x) = x. So we substitute x in for that.
```   x f'(x)  =  1                                                  eq. 6.3-13b
```
or equivalently
```             1
f'(x)  =                                                       eq. 6.3-13c
x
```
And that's the derivative of the natural log function -- 1/x. Remember it. It comes up all the time.

But your instructor might ask you to find the derivative of the natural log using limits. Recall that the derivative of a function, f(x), was defined back in unit 4 as

```                    f(x + h) - f(x)
f'(x)  =   lim                                                 eq. 6.3-14
h  > 0        h
```
And you have the limit definition of ln(x). Putting this all together, you get that if  f(x) = ln(x)  then
```                   ln(x + h) - ln(x)
f'(x) =   lim                                                  eq. 6.3-15a
h  > 0         h
```
Now substituting the limit definition for ln(x) (and using k instead of h for the limit variable in that definition) we have:
```                    (x + h)k - 1   xk - 1
-
k            k
f'(x)  =   lim                                                 eq. 6.3-15b
h  > 0           h
k  > 0
```
Here we are taking the limit as two variables, h and k, both go to zero. This looks pretty nasty, but after a little algebra and some cancellation you have
```                    1 (x + h)k - xk
f'(x)  =   lim   -                                             eq. 6.3-15c
h  > 0 k       h
k  > 0
```
Now look at the right hand factor of the product in eq. 6.3-15b. Isn't that just the limit formula for the derivative of xk? And isn't the derivative of xk equal to kxk-1? So in the limit when h goes to zero, we can substitute this expression for the right hand factor, which gives
```                    1
f'(x)  =   lim   - kxk-1                                       eq. 6.3-15d
k  > 0 k
```
Clearly the k and the 1/k cancel each other. We're left with
```   f'(x)  =   lim   xk-1                                          eq. 6.3-15e
k  > 0
```
Well that's an easy limit to take. We know exponentials are continuous, so as k goes to zero, k-1 goes to -1 and xk-1 goes to x-1. That gives
```                     1
f'(x)  =  x-1  =                                               eq. 6.3-16
x
```
So there it is, proved two different ways. The derivative of ln(x) is 1/x. And it is this property of the natural log that makes it show up so often in calculus and physics.

2) Take the derivatives of the following two expressions using the chain rule and show that they are equal.

```   f(x)  =  ln(1 + x) + ln(1 - x)
```
and
```   f(x)  =  ln(1 - x2)
```
Note that you can easily show using factoring and log identities that these two expressions are equal before you take the derivative. But see if you can take the derivative of the second without factoring the polynomial inside the log. This will exercise your skills at applying the rules for taking derivatives. Work it out and then click here and compare your answer with mine.

5) A sample of radioactive material has 6 × 1017 atoms in it. Over the course of each 72 hours, half of the atoms remaining of this material decay into a different species of atom that is not radioactive. Each decay of an atom can be detected and counted by a scintillation counter.

1. Write a function that represents the number of remaining radioactive atoms as a function of time, t, in hours.
2. Convert that function (if you haven't already) so that it uses the ex function as its exponential.
3. Write another function that expresses, as a function of t, the rate (in counts per hour) at which the scintillation counter will be counting atomic decays.
When you are done,

7) Chemists say that when a chemical system has reached a point where no further spontaneous changes are possible, it is in equilibrium. If you had, for example, a system where compound A changes into compound B, equilibrium occurs when the rate of change from A to B exactly equals the rate of change from B back to A. The system therefore is in balance.

You can find the equilibrium point of a chemical system by maximizing a quantity called entropy. The symbol commonly used for entropy is S. For chemicals in a solution, there are several variables that go into computing entropy. These are the concentration of each compound, ck, in moles per liter, the temperature above absolute zero, T, and the amount of heat energy, H, given off by the changing of one compound to another.

If the concentration of compound A is cA, then its contribution to the entropy of the system is

```  -R cA ln(cA)
```
where R is a universal physical constant.

If H is the amount of heat given off by the reaction, its contribution to entropy is

```    H

T
```
Total entropy is the sum of all the contributing terms. If you have a mole of compound A changing to a mole of compound B, for each each mole that changes from A to B you get U units of heat. Supposing we started with 1 mole per liter of A and none of B, then, as the reaction progressed, the entropy per liter would be
```                                         UcB
S  =  -R cA ln(cA)  -  R cB ln(cB)  +
T
```
Assume that temperature, T, is constant. Assume that U is constant as well. Observe that in this example, the sum of  cA + cB  is also constant at 1 mole per liter. That's because each bit of A that disappears is converted to the same amount (in moles) of B. If 1 mole per liter is that constant sum, then
```   cA  =  1 mole/liter - cB
```
Go ahead and eliminate cA from the entropy equation by substituting it with the right hand side of this equation. Entropy will be maximum when the derivative of S with respect to cB is zero:
```   dS
=  0
dcB
```
Find that derivative and set it zero. Then substitute back to find what ratio of cB:cA produces this condition. That is the equilibrium point.

You will need to use both the product rule and the chain rule to take the derivative. There will also be an important cancellation when you do the algebra.

This is a word problem with a lot of information in it, but you can solve it using the material we have covered in this section. Read it over several times and take it a little at a time. If you have studied chemistry, you will see a familiar formula emerge out of this. Get as far as you can before clicking here to see the solution.