Given: \((5^x)-(5^y)=(2^{y-1})*(5^{x-1})\)

Divide both sides by \(5^{x-1}\) to get: \(5^1 - 5^{y-x+1} = 2^{y-1}\)

Simplify: \(5 - 5^{y-x+1} = 2^{y-1}\)

OBSERVE: Notice that the right side, \(2^{y-1}\), is POSITIVE for all values of y

Since y is a positive integer, \(2^{y-1}\) can equal 1, 2, 4, 8, 16 etc (powers of 2)

So, the left side, \(5 - 5^{y-x+1}\), must be equal 1, 2, 4, 8, 16 etc (powers of 2).

Since \(5^{y-x+1}\) is always positive, we can see that \(5 - 5^{y-x+1}\) cannot be greater than 5

So, the only possible values of \(5 - 5^{y-x+1}\) are 1, 2 or 4

In other words, it must be the case that:
case a) \(5 - 5^{y-x+1} = 2^{y-1} = 1\)
case b) \(5 - 5^{y-x+1} = 2^{y-1} = 2\)
case c) \(5 - 5^{y-x+1} = 2^{y-1} = 4\)

Let's test all 3 options.

case a) \(5 - 5^{y-x+1} = 2^{y-1} = 1\)
This means y = 1 (so that the right side evaluates to 1)
The left side, \(5 - 5^{y-x+1} = 1\), when \(5^{y-x+1} = 4\). Since x and y are positive integers, it's IMPOSSIBLE for \(5^{y-x+1}\) to equal 4
So, we can eliminate case a

case b) \(5 - 5^{y-x+1} = 2^{y-1} = 2\)
This means y = 2 (so that the right side evaluates to 2)
The left side, \(5 - 5^{y-x+1} = 2\), when \(5^{y-x+1} = 3\). Since x and y are positive integers, it's IMPOSSIBLE for \(5^{y-x+1}\) to equal 3
So, we can eliminate case b

case c) \(5 - 5^{y-x+1} = 2^{y-1} = 4\)
This means y = 3 (so that the right side evaluates to 4)
The left side, \(5 - 5^{y-x+1} = 4\), when \(5^{y-x+1} = 1\).
If \(5^{y-x+1} = 1\), then \(y-x+1 = 0\)
In this case, y = 3
So, we can write: 3 - x + 1 = 0, which mean x = 4
So, the only possible solution is y = 3 and x = 4, which means xy = (4)(3) = 12

Cheers,
Brent