1)x^a=3.x^b
2)y^a=4.y^b
Should I assume that that point over there is actually "*" (multiplication)?
If so, here's my approach:
1. x^a = 3x^b means that x^a - 3x^b = 0. Now factor x^b and you get x^b[x^(a - b) - 3] = 0. Since x and b are positive, then x^b is greater than 1 (remember that any positive number raised to any positive power is greater than that number raised to 0, which is always 1). This means that, in order for the equation to be equal to zero, then x^(a - b) - 3 = 0 or x^(a - b) = 3. Since x, a and b are all positive integers, then x can only be 3, with a - b = 1. So 1 is sufficient.
2. y^a = 4y^b is equivalent to y^a - 4y^b = 0. Again, do the factoring thing and you get y^b[y^(a - b) - 4] = 0. Using the same line of thought as above you get that y^(a - b) = 4. Now, since y, a and b are all positive integers, but 4 = 2^2, you will have two options in this case:
a. y = 4 and a - b = 1
b. y = 2 and a - b = 2
Since you have no extra info, you can't pick between the two cases.
So 2 is insufficient.
So my answer would be A.
