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

log 9(x) = log 12(y) = log 16(x+y)

-> log 9(x) = log 12(x) + log 12(p) = log 16(x) + log 16(1+p), where y/x = p(say)

-> log x/log 9 = (log x + log p)/log 12
= (log x + log p)/log 12
= (log x + log (1+p))/log 16
= m(say)

Then, we must have:

log x = m*log 9, and:
log x + log p = m*log 12, and:
log x + log (1+p) = m*log 16

But, m*Log 9 + m*log 16 = m*log 144 = 2m* log 12

Accordingly, log x + log x + log (1+p) = 2(log x + log p)
or, log (1+p) = 2*log p
or, p^2 = 1+p
or, p^2 - p - 1 = 0
or, p = (1 +/- V5)/2 ........(*)

For log 9(x) and log 12(y) to be defined, both x and y must be positive.

Accordingly, the negative root of p in (*) is inadmissible, and consequently:

p = (1+V5)/2

Edited on July 13, 2008, 3:32 pm
 Posted by K Sengupta on 2008-04-06 13:10:28

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