## 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.

November 30th, 2011 at 2:31 pm

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.

November 30th, 2011 at 11:01 pm

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

December 1st, 2011 at 1:04 am

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…

December 1st, 2011 at 4:54 am

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

December 1st, 2011 at 5:29 am

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.

December 3rd, 2011 at 8:21 pm

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

December 5th, 2011 at 5:17 am

Intro: by definition x = b

^{logb(x)}= c^{logc(x)}Taking logs base b => logb(x) = logb(c

^{logc(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

x

^{3}+y^{3}−(x+y)^{3}= −3x^{2}y −3xy^{2}= −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}December 5th, 2011 at 1:24 pm

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.