Logarithms and the Area of a Triangle
Date: 6/3/96 at 16:9:38 From: Italo Wladimir de Medereiros Alves Subject: Logarithms and the area of a triangle. Is the following statement true? "If 'A' represents the area of a triangle whose sides measure a, b, and c, and if p is its semiperimeter, then log(A) = 1/2*[log(p) + log(a+b) + log(b+c) + log(a+c)]."
Date: 6/6/96 at 14:48:44 From: Doctor Brian Subject: Re: Logarithms and the area of a triangle. I would have to say no. Here's what we know. Normally, by Heron's formula, a triangle with sides a, b, and c, with semiperimeter p, has area A = sqrt[(p)(p-a)(p-b)(p-c)]. We can take the logarithm of both sides to say that log(A) = log(sqrt[(p)(p-a)(p-b)(p-c)]). Since the sqrt (square root) is really the 1/2 power of the expression on the right, we could bring that exponent outside the log so that log(A) = 1/2*(log[(p)(p-a)(p-b)(p-c)]). At this point the log of a product may be broken up as the sum of the individual logs: log(A) = 1/2*[log(p) + log (p-a) + log (p-b) + log (p-c)]. Note the similarity between this expression and your original one. The difference can be noted by seeing that (p-a) is not the same as (b+c). Since a + b + c = 2p, then b + c is = 2p - a, and so the logs of such are equal. Since p - a is smaller than 2p - a, which is equal to b + c, then log (b + c) > log (p - a). The same argument compares the other terms in the two expressions, and so your expression is greater than the one I derived. In other words, they are *not* equal. -Doctor Brian, The Math Forum Check out our web site! http://mathforum.org/dr.math/
Date: 6/3/96 at 19:19:50 From: Doctor Anthony Subject: Re: Logarithms and the area of a triangle. The formula for the area (using p = semi-perimeter) is A = sqrt[p(p-a)(p-b)(p-c)]. So A^2 = p(p-a)(p-b)(p-c) and taking natural logs 2*ln(A) = ln(p) + ln(p-a) + ln(p-b) + ln(p-c) ln(A) = (1/2)[ln(p) + ln(p-a) + ln(p-b) + ln(p-c)]. Of course you could take logs to base 10 or indeed any base. With the use of calculators there is no advantage in converting the formula to logs, but it was done in the days before calculators were available. (pre-1970s). -Doctor Anthony, The Math Forum Check out our web site! http://mathforum.org/dr.math/
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