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willbeatthegmat
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I'll assume that point = multiplication or *
5^x - 5^y= 2^(y-1)*5^(x-1) is equivalent to:
5^x - 2^(y-1)*5^(x - 1) = 5^y (i just switched the places between the two).
Factor 5^(x - 1) and you get that
5^(x - 1)*[5 - 2^(y - 1)] = 5^y.
Now, since 5^(x - 1) and 5^y are both positive (they are powers of a positive number - in this case 5), then 5 - 2^(y - 1) must be positive as well. This means in turn that 5 > 2^(y - 1) and there are only 3 cases for this one:
a. y = 1, with 2^(y - 1) = 1, and with 5 - 2^(y - 1) = 4. But this one is not right, since (5 to a power)*4 cannot equal (5 to a different power).
b. y = 2, with 2^(y - 1) = 2 and 5 - 2^(y - 1) = 3. Again, this doesn't work, since that 3 cannot "come between" powers of 5.
c. y = 3, with 2^(y - 1) = 4 and 5 - 2^(y - 1) = 1. This is the case we're looking for, since we get 5^(x - 1)*1 = 5^3. This means that x - 1 = 3, with x = 4.
This gives us xy = 12.
