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