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Find ALL positive integers x and y such that the following equation (involving Log functions BASE 2) holds:(Log(x))/(Log(y)) = Log(x/y).Prove your answer exhausts all solutions!

x=16,y=4.Haven't found any other solution but don't have a proof this is the only one yet.

QuoteOriginally posted by: phuebux=16,y=4.Haven't found any other solution but don't have a proof this is the only one yet.The heart of the problem is to prove that you've found all the solutions. It is relatively easy to come up with some solution. But is there more and how do we know it?

You can rewrite it as (log x - log y) log y = log x; (x,y) belonging to N²Then you express x and y as their integer factorization and play around with your sigmas and based on the facts the integer factorization is unique and log p / log q is not within Q if p ^ q = 1 (i mean they don't share divisors not p power q) you finally conclude that x and y must be powers of 2.So, x=2^n and y=2^m then the equation becomes (n-m)m=n; (n,m) belonging to N²thenm²-nm+n=0delta = n²-4n = k² must be a square for m to be an integerfor k=0 we get the solution from phuebuelse we get n-4<k<nk=n-3 then 2n=9, not integerk=n-2, absurdk=n-1, then 2n=1, not integerso the only solutions we get are for k=0(x,y)=(2^4,2^2)=(16,4)

Ahh, I got to the bit with n and m.

Here is my solution or rather a sketchy prove

QuoteOriginally posted by: NicolasQuantYou can rewrite it as (log x - log y) log y = log x; (x,y) belonging to N²Then you express x and y as their integer factorization and play around with your sigmas and based on the facts the integer factorization is unique and log p / log q is not within Q if p ^ q = 1 (i mean they don't share divisors not p power q) you finally conclude that x and y must be powers of 2.So, x=2^n and y=2^m then the equation becomes (n-m)m=n; (n,m) belonging to N²thenm²-nm+n=0delta = n²-4n = k² must be a square for m to be an integerfor k=0 we get the solution from phuebuelse we get n-4<k<nk=n-3 then 2n=9, not integerk=n-2, absurdk=n-1, then 2n=1, not integerso the only solutions we get are for k=0(x,y)=(2^4,2^2)=(16,4)i agree with everything except in the last step ...k=n-3 then 2n=9, not integerk=n-2, absurdk=n-1, then 2n=1, not integeryou somehow seem to assume that n is an integer. is it necessary for n to be an integer in order for x and y to be integers?