dcummins wrote:
Bunuel wrote:
Is x^5 > x^4?
Is \(x^5 > x^4\)? --> reduce by x^4: is \(x > 1\)?
(1) x^3 > −x --> \(x(x^2 + 1) > 0\). Since \(x^2 + 1\) is always positive, then the second multiple, x, must also be positive. So, this statement implies that x > 0. Not sufficient.
(2) 1/x < x --> multiply by x^2 (it has to be positive, so we can safely do that): \(x < x^3\) --> \(x(x^2 -1) > 0\) --> \((x + 1)x(x - 1) > 0\) --> \(-1 < x < 0\) or \(x > 1\). Not sufficient.
(1)+(2) Intersection of ranges from (1) and (2) is \(x > 1\). Sufficient.
Answer: C.
Hey
BunuelWhy were you able to manipulate statement 1 and 2 without knowing the sign?
Is it because the only way for x^5 to be greater than x^4 is for x to be greater than 1? So you tested the statements on this basis?
Secondly,
Why did you multiply statement 2 by x^2? Is it because we don't know the sign but we do know that x^2 > 0?
Third,
The final statements say x>0 x>1 or x>-1 from statement 2, and x^2 +1>0 and x>0 from statement 1, so combined how did we conclude the intersection was >1?
I really found testing values easier with this one, but I'm really unsure how you went about your way.
I don't understand what you mean by second multiple also
1. We can divide \(x^5 > x^4\) by x^4, because x^4 (x in even power) is non-negative, so we know the sign. This way we get simplified question: is \(x > 1\)?
2. In the firs statement we are not multiplying or dividing x^3 > −x. We are re-arranging it to get x^3 + x > 0, then factoring out x to get x(x^2 + 1) > 0. Here x is the one multiple and x^2 + 1 is another multiple. The product of two multiples to be positive both must have the same sign. Since x^2 + 1 = positive, the x, another multiple, must also be positive.
3. We are multiplying 1/x < x by x^2 to get x < x^3. Again we can do that because x^2 is non-negative.
4.
From (1)
-------------------0-----------------------From (2)
--------1---------0---------1--------------------------------Intersection:
-----1---------0---------1--------------------------------