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## Lotta logs

Posted by Chris on November 30, 2011 – 11:41 am

1. If a, b, c are positive real numbers and log(a, b) + log(b, c) + log(c, a) = 0, find the value of (log(a, b))^3 + (log(b, c))^3 + (log(c, a))^3

2. Find log(n, 1/2) * log(n-1, 1/3) * … * log(2, 1/n) in terms of n.

NB log(a, x) means log base a of x etc.

This post is under “Tom” and has 8 respond so far.

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1. 1. cazayoux Said：

Hmmm .. ain’t pretty.
for (1) I get the following

(loga b)^3 + (logb c)^3 + (logc a)^3
=
- (logc a / loga b) – (logc a / logb c)
- (loga b / logb c) – (loga b / logc a)
- (logb c / logc a) – (logb c / loga b)

I’m still looking for a way to simplify.

2. 2. Chris Said：

Hi cazayoux. It can be simplified a lot more than that.

3. 3. slavy Said：

The first part is a classical problem for 6th-7th grade (at least in Bulgaria) and meanwhile is a starting example for the theory of symmetric functions, which is very useful for computations The answer is 3, unless I did some of my standard computational mistakes

The second part can be dedicated to Gauss (if I´m not mistaken) for the way he computed the sum of the first n natural numbers many centuries ago. The answer is… wait for it…

4. 4. Chris Said：

Guess what, Slavy’s got the first one right, even though it involved an literal number (LOL).

I’m not sure about the second one – i.e. I can’t see how to get the sum of the first n natural numbers from it (but I’m no Gauss). I pretty sure that Gauss is famous for finding the sum by noticing that 1+n = 2+(n-1) = 3+(n-2) = … => sum = ((n+1)/2)n = n(n+1)/2

5. 5. slavy Said：

That’s exactly what I meant, Chris – the algorithm is important not the sum Don’t worry – this one I got it right, since there is no unnecessary arithmetics Just the answer is a much bigger hint than the one from the first part (that tells nothing about the way it is derived) and I am giving the others more time to think alone.

6. 6. Chris Said：

Hi cazayoux, you haven’t taken advantage of all the information given.

7. 7. Chris Said：

Intro: by definition x = blogb(x) = clogc(x)
Taking logs base b => logb(x) = logb(clogc(x)) = logc(x) logb(c)
=> logc(x) = logb(x) / logb(c) or logb(c) = logb(x) / logc(x)
Letting x = b => logb(c) = 1/logc(b)

For 1: Let x = loga(b) and y = logb(c), then logc(a) = −(x + y). Then we want
x3 +y3 −(x+y)3 = −3x2y −3xy2 = −3xy(x+y)
= 3 loga(b) logb(c) logc(a) = 3 loga(b) logb(c) / loga(c)
= 3 logc(b) logb(c) = 3

For 2: Using logb(1/x) = -logb(x) and logc(a) = logb(a) / logb(c)
=> the product equals
(-logb(2)) (-logb(3))…(-logb(n)) / ((logb(n)) ( logb(3))…(logb(2))) = (-1)n-1

8. 8. cazayoux Said：

Yes. I see it now.
I was playing with the fact that loga(b) = 1/logb(a), but much later in my calculations.

Chris’s substitution was the key to making it simple.
Let x = loga(b) and y = logb(c), then logc(a) = −(x + y).

Thank you. Enjoyed it.

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