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### 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/
```
Associated Topics:
High School Basic Algebra
High School Geometry
High School Triangles and Other Polygons

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