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

xwhere^{h}- 1 ln(x) = lim~~eq. 6.3-1 h~~~~> 0 h~~

That definition of the natural log function arose because of
the role it plays in
finding the derivative of
` f(x) = b ^{x}`

ln(ab) = ln(a) + ln(b) eq. 6.3-2aand

ln(bAnd you would expect it to obey those rules -- otherwise naming it the natural^{n}) = n ln(b) eq. 6.3-2b

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
`log _{10}(x)`. Recall now that for

. . . log_{10}(0.0001) = -4 log_{10}(0.001) = -3 log_{10}(0.01) = -2 log_{10}(0.1) = -1 log_{10}(1) = 0 log_{10}(10) = 1 log_{10}(100) = 2 log_{10}(1000) = 3 log_{10}(1000) = 4 . . . table 6.3-1

Figure 6.3-1 shows a plot of ` f(x) = 10 log _{10}(x)`

Logs base `10` obey the same product rule as natural logs. That is
`log _{10}(ab) = log_{10}(a) + log_{10}(b)`

You probably also learned that
` f(x) = 10 ^{x} `

logand_{10}( 10^{x}) = x

10Now 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^{log10(x)}= x

x = 10you could find a^{y}

x < 10by choosing^{yhigh}

x > 10by choosing^{ylow}

You can also readily show that `log _{10}(x)` is monotone
increasing. It follows from

I have a diversionary page that will bring back into focus what logs base
`10` are all about. It is **optional material**, and you can
get to it by

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

x = 7which follows from log and exponentiation being inverses of each other. If you take the log base^{log7(x)}eq. 6.3-3a

logDivide out_{10}(x) = log_{7}(x) × log_{10}(7) eq. 6.3-3b

logAnd in general, if you have log to the base_{10}(x) log_{7}(x) =~~eq. 6.3-3c log~~_{10}(7)

logwhere_{a}(x) log_{b}(x) =~~eq. 6.3-4 log~~_{a}(b)

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

logYou should be able to verify this yourself by raising_{b}(b) = 1 eq. 6.3-5

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:

xIf the function,^{h}- 1 ln(x) = lim~~eq. 6.3-6 h~~~~> 0 h~~

ln(x) = logAnd from equation 6.3-5, we also have_{e}(x) eq. 6.3-7

eIt turns out that the real number that satisfies this equation is irrational. To 16 places,^{h}- 1 ln(e) = lim~~= log~~_{e}(e) = 1 eq. 6.3-8 h~~> 0 h~~

(eNow we employ a trick we've used before. We substitute^{x})^{h}- 1 e^{hx}- 1 ln(e^{x}) = lim~~= lim~~~~eq. 6.3-9a h~~~~> 0 h h~~~~> 0 h~~

eBut we already know that^{k}- 1 ln(e^{x}) = lim x~~eq. 6.3-9b k~~~~> 0 k~~

eThat is because we defined^{k}- 1 lim~~= 1 eq. 6.3-9c k~~~~> 0 k~~

ln(ewhich finishes the proof and confirms that^{x}) = lim x × 1 = x eq. 6.3-9d k~~> 0~~

Now if ` ln(e) = 1`` f(x) = e ^{x} `

f'(x) = ln(e) × eThat's right.^{x}= e^{x}eq. 6.3-10

bwhich holds for any positive real number,^{x}= e^{x ln(b)}eq. 6.3-11

A few other things you should know about `e ^{x}`. Like
other exponential functions, it is defined and continuous for all real

Figure 6.3-2 shows a plot of `e ^{x}`. Observe that
this function always returns a result that is positive. Like other
exponential functions, it crosses the

The function, ` f(x) = ln(x)`

We'll do the easy one first. We know from the discussion so far that
`ln(x)` and `e ^{x}` are inverses of each other.
That means that if

eNow remember that^{f(x)}= e^{ln(x)}= x eq. 6.3-12

eBut from eq. 6.3-12, we know that^{f(x)}f'(x) = 1 eq. 6.3-13a

x f'(x) = 1 eq. 6.3-13bor equivalently

1 f'(x) =And that's the derivative of the natural log function --~~eq. 6.3-13c x~~

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) = limAnd you have the limit definition of~~eq. 6.3-14 h~~~~> 0 h~~

ln(x + h) - ln(x) f'(x) = limNow substituting the limit definition for~~eq. 6.3-15a h~~~~> 0 h~~

(x + h)Here we are taking the limit as two variables,^{k}- 1 x^{k}- 1~~-~~~~k k f'(x) = lim~~~~eq. 6.3-15b h~~~~> 0 h k~~~~> 0~~

1 (x + h)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^{k}- x^{k}f'(x) = lim -~~eq. 6.3-15c h~~~~> 0 k h k~~~~> 0~~

1 f'(x) = lim - kxClearly the^{k-1}eq. 6.3-15d k~~> 0 k~~

f'(x) = lim xWell that's an easy limit to take. We know exponentials are continuous, so as^{k-1}eq. 6.3-15e k~~> 0~~

1 f'(x) = xSo there it is, proved two different ways. The derivative of^{-1}=~~eq. 6.3-16 x~~

**1)** You already should know that
` ln(x ^{2}) = 2ln(x)`

**2)** Take the derivatives of the following two expressions using
the

f(x) = ln(1 + x) + ln(1 - x)and

f(x) = ln(1 - xNote 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^{2})

**3)** There are some functions used in advanced calculus that grow so
large so fast that it is often more convenient to deal with the natural
log of such functions. If `f(x)` is any function whose derivative
you know, write an expression in terms of `f(x)` and `f'(x)`
that gives the derivative of `ln(f(x))`. Then
Click here to see the solution.

**4)** The function,
` f(x) = e ^{-x2/2}`

**5)** A sample of radioactive material has
`6 × 10 ^{17}` atoms

- Write a function that represents the number of remaining
radioactive atoms as a function of time,
`t`, in hours. - Convert that function (if you haven't already) so that it uses
the
`e`function as its exponential.^{x} - 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.

**6)** Find the derivative of
`f(x) = log _{10}(x)`

**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, `c _{k}`,
in moles per liter,
the temperature above absolute zero,

If the concentration of compound A is `c _{A}`, then
its contribution to the entropy of the system is

-R cwhere_{A}ln(c_{A})

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

HTotal 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~~T~~

UcAssume that temperature,_{B}S = -R c_{A}ln(c_{A}) - R c_{B}ln(c_{B}) +~~T~~

cGo ahead and eliminate_{A}= 1 mole/liter - c_{B}

dSFind that derivative and set it zero. Then substitute back to find what ratio of~~= 0 dc~~_{B}

You will need to use both the

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

**8) Coached Exercise**
Demonstrate using limits and algebra only that the familiar identity
` ln(ab) = ln(a) + ln(b) ``ln(x)`. That is, you are not
allowed to appeal to the fact that `ln(x)` is the inverse of
`e ^{x}`, or to any of what you know about the derivatives
of logs and exponentials. This exercise is intended to sharpen your
ability to deal with limit problems. Since this is a coached exercise,
you may

Move on to May the Circle be Unbroken (trig functions)

email me at *hahn@netsrq.com*