Is 3^(a^2/b) < 1? (1) a < 0 (2) b < 0

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Is 3^(a^2/b) < 1?

(1) a<0

(2) b<0


When you modify the original condition and the question, they become 3^(a^2/b) < 1? --> 3^(a^2/b) < 3^0? --> a^2/b>0?. Multiply b^2 on the both equations(since b^2 is positive, even if it’s multiplied, the sign of inequality doesn’t change.) and it becomes a^2(b)>0?. There are 2 variables(a,b), which should match with the number of equations. So you need 2 equations. For 1) 1 equation, for 2) 1 equation, which is likely to make C the answer.
When 1) & 2), since a<0, it can’t be 0. Divide the both equations with a^2, they become a^2(b)<0?-->b<0?. Since 2) is b<0, it is yes and sufficient.
Therefore, the answer is C.


 For cases where we need 2 more equations, such as original conditions with “2 variables”, or “3 variables and 1 equation”, or “4 variables and 2 equations”, we have 1 equation each in both 1) and 2). Therefore, there is 70% chance that C is the answer, while E has 25% chance. These two are the majority. In case of common mistake type 3,4, the answer may be from A, B or D but there is only 5% chance. Since C is most likely to be the answer using 1) and 2) separately according to DS definition (It saves us time). Obviously there may be cases where the answer is A, B, D or E.

Bunuel ...if a=-1 and b=-9 then we get 3 as our answer and if a=-1 and b=-2 in that case we get .66
Should the answer not be E.

If a=-1 and b=-9, then \(3^{^{\frac{1}{(-9)}}} \approx 0.9\).

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