Teacher2Teacher |
Q&A #3572 |
From: George Zeliger
To: Teacher2Teacher Public Discussion
Date: 2000080305:22:46
Subject: Re: Re: Logarithms
I presume you teach logarithms to students who know pretty well properties of the exponential function, and are mature enough to understand what an inverse operation is. "Logarithming" (finding the logarithm) is inverse to the "exponentiation" (raising to the degree) by the definition of a logarithm that reads: B(ase)**L(ogarithm_base B of X)=X. When you write it on the board using usual exponential notation (which I can do using MS Word Equation Editor but cannot here) you will see -- and should very emphatically draw attention of your students to tthe fact -- that the logarithm is nothing but the exponent. The difference is that when we talk about the exponentiation, the exponent is known and it is the result of raising to that degree that we are looking for, while when we talk about the operation of finding logarithms we know the result of raising to a degree -- and the base, of course -- but do not know to what degree the base was raised. This definition must be memorized -- call it rote memorization, but without firm knowledge of it dealing with logarithms is difficult if possible at all. Once you remember it, many things become pretty simple. For example, consider the equation log(x-1)=1 assuming that the base is 10 (see Comment below). Let's assume that we found the value of x and substituted it into the equation. Then the equation turns into an equality of two numbers -- 1 and the value of log(x-1). From the definition of logarithms we know that raising the base to the degree equal to a logarithm (i.e., exponentiation of a logarithm) cancels both symbols of exponentiation and logarithming (same way as taking a square root cancels squaring; this is a genetral property of sequential application of two inverse operations). Since the symbol "log" definitely annoys us, let's get rid of it by raising the base, i.e., 10, to both parts of the equality. Obviously, we will get another equality: 10**log(x-1)=10**1, since according to our assumption log(x-1) and 1 are just the same number (there is no unknown variables anymore!) (The art of solving logarithmic equation is to transform the given equation to a form when the symbol "log" can be eliminated by raising the base to the left and the right parts of the transformed equation.) Now notice that the right part can be calculated directly, since 10**1 is just 10. At the same time, according to the definition of the logarithm, the left part merely reduces to x-1. Therefore, if we assume that we found the solution, it necessarily should satisfy the equality x-1=10. In other words, it is inevitable that x=11 -- if only the solution exists! To verify that it does exist, let's just try if the only possible applicant can really be hired. To do that, try him -- substitute the found number into the original expression and verify that it turns into a numeric equality! As a bonus, you will be able to check your calculations -- but this is an unnecessary bonus, no more than that. The true role of the substitution is verification of the assumption of the existence of a solution. George Comment: BTW, in Russia we used a special symbol for decimal logarithms, viz. "lg." Each time you see "lg" you know the base is 10. Natural logarithms were denoted "ln." "Log" was used as the generic name, but in this case the base should be shown explicitly, as a subscript to the symbol"log", e.g. log_sub2 means a binary logarithm. However, for the sake of brevity the explicit mentioning of the base sometimes is omitted in formulas that describe properties of logarithms true for any base (of course, keeping in mind that the base is the same throughout the whole formula).
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