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# Is x^5 > x^4?

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Updated on: 05 Oct 2014, 07:51
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Is x^5 > x^4?

(1) x^3 > −x
(2) 1/x < x

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Originally posted by aadikamagic on 05 Oct 2014, 07:12.
Last edited by Bunuel on 05 Oct 2014, 07:51, edited 1 time in total.
Edited the question.
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Re: Is x^5 > x^4?  [#permalink]

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05 Oct 2014, 08:00
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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.

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Re: Is x^5 > x^4?  [#permalink]

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05 Oct 2014, 09:07
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1
Thanks Bunuel. The take away for me here is that we can always divide by squares without bothering about sign changes.
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Re: Is x^5 > x^4?  [#permalink]

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06 Oct 2014, 01:29
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

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Re: Is x^5 > x^4?  [#permalink]

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06 Oct 2014, 01:55
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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

(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:
Inequality tips: tips-and-hints-for-specific-quant-topics-with-examples-172096.html#p1379270

inequalities-trick-91482.html
data-suff-inequalities-109078.html
range-for-variable-x-in-a-given-inequality-109468.html
everything-is-less-than-zero-108884.html
graphic-approach-to-problems-with-inequalities-68037.html

All DS Inequalities Problems to practice: search.php?search_id=tag&tag_id=184
All PS Inequalities Problems to practice: search.php?search_id=tag&tag_id=189

700+ Inequalities problems: inequality-and-absolute-value-questions-from-my-collection-86939.html

Hope this helps.
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Re: Is x^5 > x^4?  [#permalink]

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07 Oct 2014, 03:54
1
Is x^5 > x^4?

(1) x^3 > −x
(2) 1/x < x

It can also quickly be solved thinking about ranges.

Is $$x^5 > x^4$$?

If you think about it, the question is implicitly asking whether a value is greater than one. Indeed every other option (numbers smaller than -1; numbers between -1 and 0; numbers between 0 and 1) if plugged into the expression results in a false statement.

1) picking numbers you can figure out that:

I. the inequality is satisfied when our value is greater than zero. ( pick -1/3. (-1/3)^3 is never gonna be greater than [-(-1/3)]. Same reasoning holds for numbers <= than -1)
II. the inequality is satisfied for both proper fractions and values greater than 1.

NS

2) plugging in numbers you can figure out that the inequality is satisfied for values between both -1 and 0 and greater than 1.

NS

1+2) First statement rules out any value smaller than zero. second statement rules out any proper fraction. Our overlapping result is going to be a value grater than one, making us able to claim an answer.

C.

Hope it helps
gmat6nplus1.
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Re: Is x^5 > x^4?  [#permalink]

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07 Oct 2014, 09:31
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.

Hi Bunuel

I was able to solve the ques by plugging in values but can you explain how did x < x^3 become x(x^2-1) > 0
how did you know x^3 would be a negative quantity and took it LHS and change the inequality sign?
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Re: Is x^5 > x^4?  [#permalink]

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08 Oct 2014, 00:32
1
1
thefibonacci 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.

Hi Bunuel

I was able to solve the ques by plugging in values but can you explain how did x < x^3 become x(x^2-1) > 0
how did you know x^3 would be a negative quantity and took it LHS and change the inequality sign?

x < x^3
0 < x^3 - x
0 < x(x^2 - 1), which is the same as x(x^2 - 1) > 0 .
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Re: Is x^5 > x^4? (1) x^3 > −x (2) 1/x < x  [#permalink]

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08 Jan 2016, 10:06
2
Is x^5 > x^4?

(1) x^3 > −x
(2) $$\frac{1}{x}$$ < x

You are asked is $$x^5 > x^4$$ ---> $$x^5-x^4>0$$---> $$x^4(x-1) > 0$$ ----> as $$x^4 > 0$$ for all x ---> the question basically asks us whether x>1 ?

Per statement 1,$$x^3 > −x$$ ---> $$x^3 + x> 0$$ ---> x(x^2+1)>0 ---> x>0 (as $$x^2+1 > 0$$ for all x) but is x>1 ? no definite answer. Not sufficient.

Per statement 2, 1/x < x ---> 1/x - x<0 ---> (1-x^2)/x < 0 ---> (x^2-1)/x > 0 ---> (x+1)(x-1)/x > 0 ---> -1<x<0 or x>1 , again not sufficient to give you a definite answer.

Combining the 2 statements you get, x>1, thus giving a definite yes to the question asked.

C is thus the correct answer.

Hope this helps.
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Re: Is x^5 > x^4?  [#permalink]

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10 Jan 2016, 17:49
1
Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.

Is x^5 > x^4?

(1) x^3 > −x
(2) 1x < x

When it comes to an inequality, when a range of 1) square and 2) que includes a range of con, the con is sufficient. These two facts are important.
When you modify the original condition and the question, it becomes x^5-x^4>0?, x^4(x-1)>0?, x-1>0?, x>1?.(As x^4 is a positive number, the direction of the sign is not changed when the both equations are divided.) There is 1 variable(x), which should match with the number of equations. So you need 1 equation. For 1) 1 equation, for 2) 1 equation, which is likely to make D the answer.
For 1), x^2+1>0 is always valid from x^3+x>0, x(x^2+1)>0, which makes x>0. Since the range of que doesn’t contain the range of con, it is not sufficient.
For 2), -1<x<0, 1<x is derived from x<x^3(the direction of the sign doesn’t change when x^2 is multiplied to the both equations. When it comes to an inequality, square only matters.), 0<x^3-x, 0<x(x^2-1), 0<x(x-1)(x+1). Since the range of que doesn’t contain the range of con, it is not sufficient.
When 1) & 2), from x>1, the range of que contains the range of con, which is sufficient. Therefore the answer is C.

 For cases where we need 1 more equation, such as original conditions with “1 variable”, or “2 variables and 1 equation”, or “3 variables and 2 equations”, we have 1 equation each in both 1) and 2). Therefore, there is 59 % chance that D is the answer, while A or B has 38% chance and C or E has 3% chance. Since D is most likely to be the answer using 1) and 2) separately according to DS definition. Obviously there may be cases where the answer is A, B, C or E.
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Re: Is x^5 > x^4?  [#permalink]

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13 Jul 2017, 23:29
2
Is x^5 > x^4?

(1) x^3 > −x
(2) 1/x < x

Responding to a pm:

All the given information can be manipulated to form factors. I have discussed how to deal with multiple factors in these posts:

https://www.veritasprep.com/blog/2012/0 ... e-factors/
https://www.veritasprep.com/blog/2012/0 ... ns-part-i/
https://www.veritasprep.com/blog/2012/0 ... s-part-ii/

Is x^5 > x^4?
Is x^5 - x^4 > 0?
Is x^4(x - 1) > 0?

We ignore the even powers because x^4 will never be negative. But we need to ensure that x should not be 0 for the inequality to hold.

Is (x - 1) > 0?
Is x > 1?

Statement 1: x^3 > −x

x^3 + x > 0
x(x^2 + 1) > 0
x^2 + 1 will always be positive.
So we see that x > 0.
But we need to know whether it is greater than 1. Greater than 0 could be 0.5 or it could be 2 (or any other number). So this is not sufficient to say whether x will eb greater than 1.
Not sufficient.

(2) 1/x < x
When the sign of x is not known, we do not multiply by x. We bring it to the o there side.
(1/x) - x < 0
(1 - x^2)/x < 0
(x^2 - 1)/x > 0
(x + 1)(x - 1)/x > 0
The transition points on the number line are -1, 0, and 1. The expression will be positive when x > 1 or -1 < x < 0. This alone is also not sufficient.

Both together we know that x is positive and x > 1 or -1 < x < 0.
So x > 1 is a must.
Sufficient.

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Re: Is x^5 > x^4?  [#permalink]

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20 Sep 2017, 10:36
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.

Hey Bunuel,

How can you divide the equation by x^4 throughout ? As we don't know the value of x. We can't do that right? what if x was 0 , then we can't reduce a equation by 0

Please let me know if I am wrong. Eager to hear back from you
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Re: Is x^5 > x^4?  [#permalink]

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20 Sep 2017, 10:46
pikolo2510 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.

Hey Bunuel,

How can you divide the equation by x^4 throughout ? As we don't know the value of x. We can't do that right? what if x was 0 , then we can't reduce a equation by 0

Please let me know if I am wrong. Eager to hear back from you

If it were x^5 < x^4, then after dividing by x^4 we'd get x < 1 but x cannot be 0, so it would be x< 0 or 0 < x < 1.

x^5 > x^4 gives x > 1, so it's redundant to mention that x cannot be 0 because we got x > 1.
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Re: Is x^5 > x^4?  [#permalink]

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21 Sep 2017, 03:46
Is x^5 > x^4?

(1) x^3 > −x
(2) 1/x < x

x^5 > x^4

x^5 - x^4 > 0

x^4 ( x - 1) > 0..........x^4 is always positive........So the question boils down to:

Is x > 1?? (Note: you can divide by x^4 because it is positive and hence no sign change)

(1) x^3 > −x

x^3 + x > 0

x ( x^2 + 1) > 0...........( x^2 + 1) is always positive.......then x cant take any positive number

Let x = 1/2........... Answer is No

Let x =2 ...............Answer is Yes

Insufficient

(2) 1/x < x

To be safe, test 1/2, -1/2, 2, -2.......... (Note: -2 & 1/2 will yield Invalid results so they are out.)

Let x = -1/2.........Answer is No

Let x = 2 ............Answer is Yes

Insufficient

Combine 1 & 2

The intersection happens when x > 1

Sufficient

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27 Dec 2017, 01:40
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.

Can we just reduce the equation by $$x^4$$, when we do not the sign of the variables? Will that not be wrong?
If it is right to reduce by $$x^4$$ in this question, how do we assess this for other questions? What do we base our judgment on to divide by variables?

Or is you reduced here because $$x^5> x^4$$ could only hold if $$x>1$$?
Separate from the question above

if we manipulated the stem this way would it be right?
$$x^5 - x^4>0$$
$$x^4(x - 1)>0$$

Therefore
either $$x^4>0 --> x > 0$$

$$Or x >1$$

Kindly explain.
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Re: Is x^5 > x^4?  [#permalink]

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27 Dec 2017, 02:15
mtk10 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.

Can we just reduce the equation by $$x^4$$, when we do not the sign of the variables? Will that not be wrong?
If it is right to reduce by $$x^4$$ in this question, how do we assess this for other questions? What do we base our judgment on to divide by variables?

Or is you reduced here because $$x^5> x^4$$ could only hold if $$x>1$$?
Separate from the question above

if we manipulated the stem this way would it be right?
$$x^5 - x^4>0$$
$$x^4(x - 1)>0$$

Therefore
either $$x^4>0 --> x > 0$$

$$Or x >1$$

Kindly explain.

When dividing by negative value we should flip the sign of the inequality, when dividing by positive value we should keep the sign of the inequality. x^4 cannot be negative, so we can divide and write x > 1.

If it were x^5 < x^4, then after we divide by x^4, we'd get x < 1 but here we should mention that x ≠ 0 because if x = 0, then x^5 = x^4 = 0.
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Re: Is x^5 > x^4?  [#permalink]

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27 Dec 2017, 09:36
1
Is x^5 > x^4?

(1) x^3 > −x
(2) 1/x < x

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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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.

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?

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Re: Is x^5 > x^4?  [#permalink]

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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.

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?

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.

9. Inequalities

For more check Ultimate GMAT Quantitative Megathread

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GMAT 4: 650 Q44 V36
GMAT 5: 650 Q48 V31
GMAT 6: 600 Q38 V35
GMAT 7: 710 Q47 V41
GPA: 3
WE: Management Consulting (Consulting)
Re: Is x^5 > x^4?  [#permalink]

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20 Aug 2019, 14:29
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.

Hey Bunuel

Why 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
_________________
Here's how I went from 430 to 710, and how you can do it yourself:
Re: Is x^5 > x^4?   [#permalink] 20 Aug 2019, 14:29

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