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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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 solution | Comment 1 of 10

Let the three equal logarithms all be equal to a number we'll call n.

Translating the equations into exponential form, we get:
x=9^n
y=12^n
x+y=16^n

y/x = 12^n/9^n = (4/3)^n

(x+y)/y = 16^n/12^n = (4/3)^n

So (x+y)/y = y/x.

Then y^2 - xy - x^2 = 0

y = (x +/- sqrt(5*x^2))/2

y/x = (1 +/- sqrt(5))/2

Since y > x,  y/x = (1+sqrt(5))/2, the golden ratio.

The numeric way would be via a computer program:

DEFDBL A-Z
n = 1.5
DO
pn = n
n = LOG(9 ^ n + 12 ^ n) / LOG(16)
LOOP UNTIL pn = n
PRINT n, 12 ^ n / 9 ^ n, 9 ^ n / 12 ^ n, 9 ^ n, 12 ^ n, 16 ^ n

The second number printed in the results below is the desired answer, the first being n, and then the second line has x, y and x+y:

` 1.672720934462332           1.618033988749894           .6180339887498949 39.46221170149797           63.85119980426752           103.3134115057655`

 Posted by Charlie on 2004-12-08 15:29:24

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