M7MBA wrote:If a > 0, is $$t^a>w^a?$$

(1) t > w
(2) t = 2w

The OA is C.

Why is C the correct answer? I don't have this DS question very clear. I need some help.

(1) t > w

Case 1: Say a = 2, t = 1/2 and w = 1/4, then we have t > w, and t^a > w^a => (1/2)^2 > (1/4)^2 => 1/4 > 1/16. The answer is Yes.
Case 2: Say a = 2, t = -1/4 and w = -1/2, then we have t > w, and t^a > w^a => (-1/4)^2 ? (-1/2)^2 => 1/16 < 1/4. The answer is No. No unique answer. Insufficient.

(2) t = 2w

Plugging-in the value of t = 2w in t^a > w^a, we have (2w)^a > w^a => 2^a*w^a > w^a => w^a(2^a - 1) > 0.

If w = 0, then the answer is No. Else, we have w^a(2^a - 1) > 0 => (2^a - 1) > 0 => 2^a > 1

Since a > 0, we have 2^a > 1. The answer is Yes. No unique answer.

(1) and (2) togther

If w = 0, then from (2), we have t = 2w => t = 2*0 = 0, but from (1), we have t > w, thus, w ≠ 0, so we have 2^a > 1. The answer is Yes. A unique answer. Sufficient

The correct answer: C

Hope this helps!

-Jay
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