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### Logarithms

```
Date: 01/25/98 at 04:11:39
From: Eveline
Subject: Logarithm

Why are there so many rules? They are so difficult to do.
```

```
Date: 01/28/98 at 12:42:09
From: Doctor Joe
Subject: Re: Logarithm

Dear Eveline,

Log is a very useful mathematical tool, widely used in all areas of
Science: Chemistry, Physics, Astronomy, Biology, and even in Social
Sciences and English Literature and Linguistic Studies.

The main objective in using Log dates back to its invention by John
Napier, an astronomer (stargazer). Napier had lots of tedious work
multiplying very large numbers (which represent the distance between
two stars/galaxies). He figured out that by reducing these large
numbers to smaller ones, not only was effort saved (because data used
for computation are smaller) but also the operaton of addition could

I shall explain why this is so.

First let us make clear what Log means.

We start by representing a number, say 10, in the following way:

10 = 10^1 (read as 10 to the power of 1)

Next we look at the number 100.  Can we write it as a power of 10?
Yes!  A bit of trial and error would lead us to 100 = 10^2. So we see
that the index (power) has increased from 1 to 2.

If we could represent 10 by the number 1 and 100 by the number 2, then
we would have smaller numbers to play with. Computationally speaking,
we might be better off.

Let's try multiplying 10 by 100. Our usual mulltiplication tells us:
10 x 100 = 1000, but 1000 is 10^3. Note that 1 + 2 = 3. The powers add
up when the original numbers they represent are multiplied together!

But this is not surprising if we think further. 10 x 100 means one
copy of 10 multiplied by 2 copies of 10. We know that that sums to 3
copies of 10, i.e. 10^3 = 1000.

In general, we see that if a is not zero,

a^m x a^n = a^(m+n)

for all integers m and n.

Furthermore, for certain good reasons of "continuity" (I shall not
deal with this in detail; just bear in mind that this is often the
convention), the property extends to generally all values of m and n:

For a positive a, a^m x a^n = a^(m+n) for all real values of m and n.

Now note that we want to represent a number (in the first example, it
was 10; in the second example, it was 100) as a power of 10. But we
know that not every number is a "nice" power of 10 (meaning, not every
number is an integer power of 10. For example the number 5 does not
end in any zeros, so it cannot be a perfect power of 10).

Here in our case, the art of plotting curves helps (this also ties in
with the concept of "continuity"). First look at the following table
of values:

If x takes on the possible values:  1, 2, 3, 4, then y = 10^x will
take on the corresponding values of 10, 100, 1000, 10000

We obtain:

x |   1   |   2   |   3   |   4   |
------------------------------------
y |  10   |  100  | 1000  | 10000 |

If we plot these points out, we have (not drawn to scale)

y

|10000                      *
|
|
|
|
|
|
|
|
|
|
|
|                    *
|
|             *
|      *
-------------------------------------------
0      1      2      3      4

We see that a smooth (and very steep) curve can be drawn to join the
points. In this way, we will be able to tell (up to a few decimal
places) what value of x would give 5 as the y-value. That value, say
p, will satisfy the property 10^p = 5. We will have sucessfully
represented 5 as a power of 10 (although this time p is no longer an
integer).

If we draw the graph using a very accurate computer, we will be able
to present any positive number as a power of 10.

Instead of having to mention when 10 is raised to "such and such a
power" to obtain another number, we invent a mathematical symbol to
represent that power.

Suppose a (not equal to 1) is a positive number, and x is a number
(obtained from the graph) such that

10^x = a

Then, we write x as Log a.

Remember that Log (something) stands for the index of 10; i.e.

Log (x)
10         =  x

In case a = 1, we set Log 1 = 0.  (Why?)

To explain this special case, we need to look at yet another law of
indices (the first law being:  a^m x a^n = a^(m+n)).

Rule 2:  For a positive a,  a^m / a^n = a^(m-n)  for all m,n real.

This law is quite natural and evident in the following example:

For the case when m,n are positive integers,

a^m   a x a x a x ...x a x a x...x a
--- = ------------------------------
a^n   a x a x a x ...x a

There are m a's on top and n a's below. So, the usual cancellation law
sets in and one starts cancelling the n a's from the m a's (assuming
that the top has more than the bottom; but it does not matter). Thus,
it is natural that there are (m-n) a's remaining.

a^m / a^n = a^(m-n).

......

Finally, let's look at these rules in the light of Log:

Rule 1: 10^m x 10^n = 10^(m+n) for all real values m and n.

Writing this statement in the language of Log:

First of all, m = Log(10^m).  You can say also that n = Log(10^n).

Next, we write x in place of 10^m and y in place of 10^n.

This rule of indices says that

m+n = Log(10^(m+n))

= Log(10^m x 10^n)

= Log(xy)

On our lefthand side, we have m + n = (Log x) + (Log y). (Why?  Look
at the definition of x and y.) Thus, we arrive at the fact that Log
(xy) = (Log x) + (Log y).

Look at the implication of this statement.

Suppose that x and y are large positive numbers. By finding the Log
of these numbers, we are reducing the size tremendously (that's why I
say that the graph we just plotted is very steep... it is sometimes
called an exponential curve). Second, to compute the Log of the
product of x and y, we need only compute the sum (addition) of the
smaller quantities Log x and Log y. This is what I intend to
illustrate (from the example of John Napier).

Of course, in general we do not need to restrict ourselves to using
the number 10. We can use any positive number (but not 0 or negative
numbers, for continuity's sake), say a instead of 10.

Then, generally for a (not equal to 1),

Log_a (x) = y means that a^y = x.

The lefthand quantity Log_a (x) is read as the logarithm of x,
based a.

(In the previous example, Log is just the same as Log_10.)

So, the first law of indices may be restated as:

Rule 1: Log_a (xy) = Log_a (x) + Log_a (y).

Logically, the representation of the second law of indices can be
proven:

Rule 2: Log_a (x/y) = Log_a (x) - Log_a (y).

In addition, for a positive real number a and an integer power p, and
another positive number b,

if y = a^p,  then log_b (y) = log_b (a^p).
= log_b (a x a x ... x a)

(p copies of a's)

It follows from repeated usage of rule 2 that

log_b (y) = log_b (a) + log_b (a) + ... + loh_b(a)

(p summands of log_b (a))

Thus, we have the third law:

log_b (a^p) = p log_b (a).

(as if the power were being brought down!)

This rule extends to all real values of p. (again by the continuity

I hope you see why Log is useful.

P.S. Bear in mind that Log just helps us represent positive numbers as
powers of a fixed positive number (called the base, in the above
example, it is denoted by a). In any problem where you can't decide
which rule to use, rewrite the statement involving Log's to statements
involving the powers of the base.  Then, it usually becomes clearer

-Doctor Joe,  The Math Forum
Check out our web site!  http://mathforum.org/dr.math/
```

```
Date: 02/05/98 at 03:13:50
From: Eveline
Subject: Logarithm

4^x+3 = 7^x-1
```

```

Date: 02/05/98 at 05:31:31
From: Doctor Mitteldorf
Subject: Re: Logarithm

Dear Eveline,

By graphing, or by trial-and-error on a calculator, you can come close
to the answer, about 1.1. For this type of equation, there is no
algebraic solution that will turn your equation into a formula. The
best you can do is to follow a procedure that gets you closer and

For example, take your equation, add 1 to both sides, then take the
natural log of both sides:

ln(4^x+4) = ln(7^x) = x*ln(7)

So ln(4^x+4)/ln(7) = x

In other words, when you find the right x, you will be able to put it
into the formula ln(4^x+4)/ln(7) and get back out the number you
started with, x. What if you put in a number that's a little off from
the right answer? What usually happens in this situation is one of two
things: either applying the formula gets you closer to your answer or
gets you farther away.

Constructing the formula for x in terms of x from your equation wasn't
hard, but it takes some experience to assure that the formula is one
of the first type, that gets you closer rather than further. Once
you've done that, you can "iterate". That means program your
calculator or your computer (or you can do it by hand on a calculator,
with a lot of keystrokes carefully entered) to repeat the function
over and over. Start with x=1 going in, and find x=1.0686... coming
out. Now put that x in, and get out 1.093. Keep going. If you're using
a programmable calculator, it's easy to just press the button for your
program 8 or 10 times. You'll see the number change more and more
slowly, converging to 1.108859...

-Doctor Mitteldorf,  The Math Forum
Check out our web site!  http://mathforum.org/dr.math/
```
Associated Topics:
High School Logs

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