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/
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