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Re: Is x^5 > x^4?
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27 Dec 2017, 09:36
aadikamagic wrote: Is x^5 > x^4?
(1) x^3 > −x (2) 1/x < x The answer is C From 1 we can infer that the value of x is greater than 0 but we can not be sure that it is not a fraction . suppose x=0.2 then 0.008>0.2 but 0.2^5<0.4 and if we take integer then we will get x^5 > x^4 hence two cases insufficient . Multiply both the sides of the statement 2 we get x<x^3 or x^3x>0 or x*(x+1)*(x1)>0 there are two possible ranges for the equation 1<x<1 or x>1 hence insufficient . Together they are sufficient as x>1
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Re: Is x^5 > x^4?
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16 Jun 2018, 17:04
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. Hi Bunuel, From 2 it seems we are able to derive that x cannot equal 0 since 1/x < x and 1< x^2. Why then do we not just do as follows: 1 < x^2 0 < x^2  1 (x1)(x+1) > 0 So, x>1 or x<1. Also, keeping in mind that: 1/x < x , then x>0. Then, from the above x>1. My tangential questions here would be how did you know to multiply both sides by x^2? Thanks in advance.



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Re: Is x^5 > x^4?
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16 Jun 2018, 22:05
energy2pe 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. Hi Bunuel, From 2 it seems we are able to derive that x cannot equal 0 since 1/x < x and 1< x^2. Why then do we not just do as follows: 1 < x^2 0 < x^2  1 (x1)(x+1) > 0 So, x>1 or x<1. Also, keeping in mind that: 1/x < x , then x>0. Then, from the above x>1. My tangential questions here would be how did you know to multiply both sides by x^2? Thanks in advance. Yes, since x is in the denominator, then it cannot be 0 but we still don't know its sign. If x > 0, then when crossmultiplying 1/x < x we'd get 1 < x^2. If x < 0, then when crossmultiplying 1/x < x we'd get 1 > x^2 (flip the sign when multiplying by negative value). Next, when we multiply an inequality by a variable in even power (x^2, x^4, ...), we don't have to worry about its sing because a number in even power cannot be negative. 9. Inequalities For more check Ultimate GMAT Quantitative Megathread
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Re: Is x^5 > x^4?
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09 Aug 2018, 10:37
Bunuel wrote: shreyagmat wrote: Bunuel, can you please explain this :"x < x^3 > x(x^2 1) > 0 > (x + 1)x(x  1) > 0 > 1 < x < 0 or x > 1. Not sufficient."
" i understand that " (x + 1)x(x  1) > 0" and then the three roots are = 1,1,0. Since the inequality is ">", x is " outside these values. But how did we come to "1 < x < 0 or x > 1."
I came to : x(x+1)(x1)>0 either x>0 or (x+1)(x1)>0
and then im lost. Please Help (x + 1)x(x  1) > 0 > the "roots", in ascending order, are 1, 0 and 1, this gives us 4 ranges: x < 1; 1 < x < 0; 0 < x < 1; x > 1. Next, test some extreme value for x: if x is some large enough number, say 10, then all three multiples will be positive which gives the positive result for the whole expression, so when x>1, the expression is positive. Now the trick: as in the 4th range expression is positive, then in the 3rd it'll be negative, in the 2nd it'll be positive and finally in the 1st it'll be negative:  +  + . So, the ranges when the expression is positive are: 1 < x < 0 and x > 1. Theory on Inequalities: Solving Quadratic Inequalities  Graphic Approach: http://gmatclub.com/forum/solvingquadr ... 70528.htmlInequality tips: http://gmatclub.com/forum/tipsandhint ... l#p1379270http://gmatclub.com/forum/inequalitiestrick91482.htmlhttp://gmatclub.com/forum/datasuffine ... 09078.htmlhttp://gmatclub.com/forum/rangeforvar ... 09468.htmlhttp://gmatclub.com/forum/everythingis ... 08884.htmlhttp://gmatclub.com/forum/graphicappro ... 68037.htmlAll DS Inequalities Problems to practice: http://gmatclub.com/forum/search.php?se ... tag_id=184All PS Inequalities Problems to practice: http://gmatclub.com/forum/search.php?se ... tag_id=189700+ Inequalities problems: http://gmatclub.com/forum/inequalityan ... 86939.htmlHope this helps. Dear Bunuel, Thank you for the amazing explanation. But I'm actually confused on this step: (x + 1)x(x  1) > 0 > the "roots", in ascending order, are 1, 0 and 1, this gives us 4 ranges: x < 1; 1 < x < 0; 0 < x < 1; x > 1. Kindly explain how does this provide us with the afore mentioned values: I'm actually stuck here: (x + 1)x(x  1) > 0; So, (x+1)>0 x > 1; x>0; (x1) > 0 x > 1; Kindly help me with this; Im unable to reach the 4 ranges you mentioned.




Re: Is x^5 > x^4? &nbs
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