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Problem Of The Week 52

with 5 comments


Suppose that and are positive numbers for which

\log_{9}p =\log_{12}q=\log_{16}(p+q),

what is the value of ?

Written by Implex

October 9, 2008 at 8:46 pm

5 Responses

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  1. Let,

    log(9) p = log(12) q = log(16) p+q = k
    Hence,
    [log p]/2 = k log3 ——1
    [log(p+q)]/2 = k log4——-2
    log q = k log12——–3

    Adding 1 and 2-
    [log (p^2 + pq) ]/ 2 = k log12

    p^2 + pq = q^2
    1+ q/p = (q/p)^2

    x^2 – x – 1 = 0
    (x-1/2)^2 = 5/4
    x = [1+-sqrt(5)]/2
    Since x=q/p>0

    x= [1+sqrt5]/2

    Amit

    October 10, 2008 at 7:59 pm

  2. Let the three expressions be equal to k.
    Now it’s easy 9^k+12^k=16^k
    We can easily see that 2 satisfies.
    So q/p=144/81=16/9

    Celebrating Life

    October 13, 2008 at 8:13 am

  3. Oops Monday morning blues ..Just ignore my above post:O

    Celebrating Life

    October 13, 2008 at 8:18 am

  4. Ah, taking 9^k+12^k=16^k
    Then 1+(4/3)^k=(4/3)^2k
    Now the equation is t^2-t-1=0 where t=(4/3)^k
    now t>0 so t=(1+sqrt(5))/2..Now q/p=(12/9)^k=t

    Celebrating Life

    October 13, 2008 at 8:27 am

  5. since,
    9^k+12^k=16^k
    Then 1+(4/3)^k=(4/3)^2k
    Now the equation is t^2-t-1=0 where t=(4/3)^k

    => t = {1+ sqrt(5)} / 2
    and p/q = 9^k / 12^k = (3/4)^k = (4/3)^-k
    hence 2/(1+(sqrt5)) ans.

    yodha

    October 16, 2008 at 12:25 pm


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