If a x =b and b y =a then xy=?

Substitute the second equation into the first. Since b = a^x, we get a = b^y = (a^x)^y. By the power-of-a-power rule, (a^x)^y = a^{xy}, so the relation becomes a = a^{xy}.

For a general base a (not 0, 1 or -1), two equal powers of the same base must have equal exponents, so xy = 1. Option A is correct.

Why the others fail:

• B) -2 and D) 2: These would need a^{-2} = a or a^{2} = a, which holds only for special values of a, not in general.
• C) -1: This needs a^{-1} = a, i.e. a^2 = 1, which holds only when a = ±1. Those are degenerate bases where the relation gives no real information, so they cannot be the intended general answer.

Concept: laws of indices and comparing exponents on equal bases. It is basic algebra, but the same logic underlies logarithms and exponential growth, which appear in Quant and in Mains data interpretation.

Trap: Aspirants either plug in random numbers without checking the degenerate cases, or write (a^x)^y as a^{x+y}. Use (a^x)^y = a^{xy} and then equate exponents.