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 Multi-Logarithms (Posted on 2004-12-08)
If log9(x) = log12(y) = log16(x+y), then find y/x.

 See The Solution Submitted by SilverKnight Rating: 3.4000 (5 votes)

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 Puzzle Solution (Method II) | Comment 9 of 10 |
(In reply to Puzzle Solution by K Sengupta)

Let, log_9(x) = log_12(y) = log_16(x+y)= m(say)
-> log_10(x)/log_10(9) = log_10(y)/log_10(12) = log_10(x+y)/log_10(16) = m
-> log_10(x)= m*log_10(9), log_10(y) = m* log_10(12), and log_10(x+y) = m*log_10(16)
.......(*)

But, we observe that:

log_10(9) + log_10(16) = log_10(144) = log_10(12^2)= 2* log_10(12)
-> m*log_10(9) + m*log_10(16) = 2*m*log_10(12)
-> log_10(x) + log_10(x+y) = 2*log_10(y)
-> x(x+y) = y^2
-> y^2 -xy - x^2 = 0
-> (y/x)^2 - 2*(y/x) - 1 = 0
-> y/x = (1+/- V5)/2

But, if y/x = (1-V5)/2 ~ 0.618, then precisely one of x or y must be negative. In that situtation, log_9(x) would be undefined if x is negative, while log_12(y) would be undefined if y is negative.

Consequently, y/x = (1+ V5)/2

 Posted by K Sengupta on 2008-02-28 05:39:49

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