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Is x^5 > x^4?

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Re: Is x^5 > x^4?  [#permalink]

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New post 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^3-x>0 or x*(x+1)*(x-1)>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?  [#permalink]

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New post 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
(x-1)(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?  [#permalink]

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New post 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
(x-1)(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 cross-multiplying 1/x < x we'd get 1 < x^2.
If x < 0, then when cross-multiplying 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.

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Re: Is x^5 > x^4?  [#permalink]

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New post 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)(x-1)>0
either x>0 or (x+1)(x-1)>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/solving-quadr ... 70528.html
Inequality tips: http://gmatclub.com/forum/tips-and-hint ... l#p1379270

http://gmatclub.com/forum/inequalities-trick-91482.html
http://gmatclub.com/forum/data-suff-ine ... 09078.html
http://gmatclub.com/forum/range-for-var ... 09468.html
http://gmatclub.com/forum/everything-is ... 08884.html
http://gmatclub.com/forum/graphic-appro ... 68037.html

All DS Inequalities Problems to practice: http://gmatclub.com/forum/search.php?se ... tag_id=184
All PS Inequalities Problems to practice: http://gmatclub.com/forum/search.php?se ... tag_id=189

700+ Inequalities problems: http://gmatclub.com/forum/inequality-an ... 86939.html

Hope 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;
(x-1) > 0
x > 1;

Kindly help me with this; Im unable to reach the 4 ranges you mentioned.
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Re: Is x^5 > x^4? &nbs [#permalink] 09 Aug 2018, 10:37

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