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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 
be used instead of multiplication.

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 
business).

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 
where you are heading.

-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 
closer to the answer.

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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