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Q&A #3572

Teachers' Lounge Discussion: Logarithmic functions

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From: George Zeliger <zeliger.g@asqnet.org>
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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