I need to find a lower limit of 2^x-3^y with x, y Integers and
2^x>3^y. 2^x-3^y=0 for y=x*Log(2)/Log(3) (which is an irrational
number and therefor always bigger than it's integer part) so the
minimum should be reached at 2^x-3^IntegerPart(x*Log(2)/Log(3)). The
problem is therefor equal to the question of how small
x*Log(2)/Log(3)-IntegerPart(x*Log(2)/Log(3)) can get up to a given x,
or how close does the graph of the straight line with slope
Log(2)/Log(3) get to a tuple of integers (up to a certain limit for
the coordinates).
The limit of 2^x-3^y should at least be bigger than or equal to
3^y/2^y-1. Looks rather trivial, especially since 2^x-3^y seems to
explode while the other expression grows comparably slowly.

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Clarification of Question by gatpog-ga on 14 Dec 2005 15:16 PST

x and y should be natural numbers, not integers

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Clarification of Question by gatpog-ga on 17 Dec 2005 02:07 PST

That's not what I meant. The limit should be defined by either x or y.
The decisive part of my question is 2^x-3^y >= 3^y/2^y-1.
Eg: y=5
3^5=243
For x=Int(y*Log(3)/Log(2))+1=Int(7,92...)+1=8 the difference 2^x-3^y
reaches a minimum (note that 2^x > 3^y): 2^8-3^5=256-243=13
13 is actually bigger than (3^2)^5-1=243/32-1=6,59375
The question is how close does 2^x get to 3^y for a given x or y.

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Clarification of Question by gatpog-ga on 17 Dec 2005 14:36 PST

"Try x = 4701, y = 2966.  This seems to be the best solution for x < 5000."
I don't know what you mean. 2^4701-3^2966 has more than 1400 digits.
(3/2)^2966-1 only has about 500, so this pair also fulfills the
uniquation 2^x-3^y >= (3/2)^y-1.

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Clarification of Question by gatpog-ga on 19 Dec 2005 02:25 PST

Thank you for your clarification emoll-ga. That's almost what I'm
searching for. I don't need a function that gives me the pair of
integers that causes the minimum, but the minumum itself. So f(5000)
should be 0.0007715. But I don't even need an exact result, I just
need a lower limit for it. For example f(x) =1^(-x) (though I wouldn't
know how to prove that this is a lower limit for all natural numbers
x, and the prove is what I need) which would give 1^(-5000) in your
example which is certainly smaller than the exact value 0.0007715. I
could use this function in my original term 2^x-3^y >=
2^x-3^(x*Log(2)/Log(3)-f(x)) but the result of this has to be bigger
than (3/2)^y-1. Again, I need a prove that the unequation 2^x-3^y >=
(3/2)^y-1 is true for all natural numbers x and y.

Your clarification has confused me.  You clarified that x and y are
natural numbers not integers.  So how exactly do integers figure into
your question?

Hello, gatpog.  If I am interpreting your question as you intended,
the answer is pretty straightforward: if x and y are natural numbers,
then both 2^x and 3^y are also natural numbers.  The theoretical lower
limit for the difference between two unequal natural numbers is 1, and
for 2^x-3^y this limit is achieved by taking x = 2, y = 1, which gives
2^x-3^y = 2^2-3^1 = 4-3 = 1.

If you did mean x and y are integers (as opposed to natural numbers),
I still believe one could make 2^x-3^y smaller than any specified
positive number by taking appropriate "large negative" values for x
and y.  For example, I took epsilon = .001 and with minimal effort
found 2^(-10)-3^(-7) = .0005 < epsilon.

I guess the question is intended to ask what is 
   inf{ 2^x - 3^y | x,y integers -> positive infinity }.

Since 2^x-3^y is an integer this is the same as asking what is the smallest
number c such that there are infinitely many pairs of positive integers
(x,y) with 2^x-3^y=c.

The answer is there is no such c. So the answer to your original question:
the "lower limit of 2^x-3^y for integers" is positive infinity.

In fact, Polya proved in 1918 that for a fixed integer c, the Dopphantine
equation
   a^x-b^y=c in positive integers a,b,x,y
has only finitely many solutions.

The Catalan conjecture states that for the case c=1 the only solution
with x,y>1 is 3^2-2^3=1. This "had defied all attempts to prove it for
more than 150 years" (MathWorld) and was finally (believed to be soved)
by Mihailescu in 2002, following steps of many other mathematicians.
Also related is Pillai's conjecture. You can find many related articles 
in Google.

Try x = 4701, y = 2966.  This seems to be the best solution for x < 5000.

I will try to explain my comments clearly.  I apologize if they do not
satisfactorily answer the precise issue you mean to address.

If x and y are both restricted to being natural numbers, and we
require 2^x > 3^y, the single ordered pair (x,y) that minimizes
2^x-3^y is (2,1).
For a given natural number x (>1), the natural number y that minimizes
2^x-3^y (still requiring 2^x > 3^y) is given by y = int(x*log2/log3).
For a given natural number y, the natural number x that minimizes
2^x-3^y (still requiring 2^x > 3^y)
is given by x = int(y*log3/log2)+1.

Where did (4701,2966) come from?  Within your original post, you said:
   "The problem is therefor equal to the question of how small
  x*Log(2)/Log(3)-IntegerPart(x*Log(2)/Log(3)) can get up to a given x."

If x is restricted to natural numbers less than 5000, the expression
x*Log(2)/Log(3)-IntegerPart(x*Log(2)/Log(3)) has a minimum where x =
4701 (in which case the expression evaluates to approximately
0.0007715).  The corresponding y (i.e., the maximum integral y such
that 2^4701 > 3^y) is 2966.

If you are looking for a function that would, say, give f(5000) =
4701; f(1000) = 485; f(100) = 65; etc., where the returned value is
that (natural number) x less than or equal to the input value such
that x*Log(2)/Log(3)-IntegerPart(x*Log(2)/Log(3)) is minimized, that
could be more difficult.  But a similar function that would instead
seek to minimize 2^x-3^(int(x*log2/log3)) would, for an input value of
2 or greater, always return 2.