If the bases were not prime numbers but still equal to one another, could the problem be solved with the same methodology, ie setting the exponents equal to one another?
If (22x+1)(32y-1) = 8x27y, then x + y =
-3
-1
0
1
3
Let's rewrite the right side of the equation in base 2 and base 3: (22x+1)(32y-1) = (23)x(33)y. This can be rewritten as: (22x+1)(32y-1) = 23x33y
Since both bases on either side of the equation are prime, we can set the exponents of each respective base equal to one another:
2x + 1 = 3x, so x = 1
2y - 1 = 3y, so y = -1
Therefore, x + y = 1 + (-1) = 0.
The correct answer is C.