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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… :P

  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 :P 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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