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# Is x > 0?

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Updated on: 12 Apr 2013, 02:24
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35% (medium)

Question Stats:

66% (01:21) correct 34% (01:26) wrong based on 175 sessions

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Is x > 0?

(1) x^6 > x^7
(2) x^7 > x^8

Originally posted by aDAMS on 11 Apr 2013, 18:16.
Last edited by Bunuel on 12 Apr 2013, 02:24, edited 1 time in total.
RENAMED THE TOPIC.
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Re: DS question on Exponents  [#permalink]

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11 Apr 2013, 18:49
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2
Stmt 1 is Not Sufficient

x^6 > x^7 means x < 1, so x > 0 when 0< x < 1 and when x < 0 so x < 0

Stmt 2 is Sufficient

x^7 > x ^8 means 0<x<1 so x > 0

A variable with smaller power is more than a variable with larger power iff the variable is in the range 0 < x < 1

//kudos please, if the above explanation is good.
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Re: DS question on Exponents  [#permalink]

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11 Apr 2013, 23:09
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2
Q. Is x > 0?

1. x^6>x^7
2. x^7>x^8

Any number with an even exponent is $$>0$$, but I wanna give you an alternative solution:
1. $$x^6>x^7$$ so $$0>x^7-x^6, 0>x^6(x-1)$$ at this point we can divide by x^6 because we know that does not equal 0(otherwise x^6=x^7 and not >) $$0>x-1$$ and finally $$1>x$$. Is x positive? it could be(0,5) or not (-1)

2.$$x^7>x^8$$ becomes $$0>x^6(x^2-x)$$ divide $$0>x^2-x$$ so $$x(x-1)<0$$ and $$0<x<1$$. Is x positive? yes
B
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Re: Is x > 0?  [#permalink]

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12 Apr 2013, 02:38
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Re: Is x > 0?  [#permalink]

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12 Apr 2013, 04:35
Felt the following would be a simpler explanation

The two option would go as follows

1. x^6 > x^7
This option is only possible either when 1>x>0 or x<0
Like if x is <1 but >0 then the values will keep on decreasing with every increasing exponent
and if x<0 then the even powers will be positive and odd ones would be -ve therefore irrespective of the value of |x| the even exponents will always be greater

2. x^7 > x^8
This is only possible when 1>x>0, as the situation would be as in case 1 but when X<0 the even exponents will always be greater

Only 2 is sufficient but 1 is not B
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Re: Is x > 0?  [#permalink]

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12 Apr 2013, 06:59
Sorry this is probably as silly as it gets but i need help in understanding the below

In x^7 > x^8, say we divide by x^6,
x> x^2
0>x(x-1)
Now 0>x and 0>(x-1)
for 0>x-1 ==> x <1, till here i am fine
but how do we get x>0 to make 0<x<1 ??
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Re: Is x > 0?  [#permalink]

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12 Apr 2013, 07:04
summer101 wrote:
Sorry this is probably as silly as it gets but i need help in understanding the below

In x^7 > x^8, say we divide by x^6,
x> x^2
0>x(x-1)
Now 0>x and 0>(x-1)
for 0>x-1 ==> x <1, till here i am fine
but how do we get x>0 to make 0<x<1 ??

x*(x-1)<0 implies

either x<0 and x-1 > 0
or x>0 and x-1<0

first is impossible and hence from second we have 0<x<1
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Re: Is x > 0?  [#permalink]

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12 Apr 2013, 07:56
summer101 wrote:
Sorry this is probably as silly as it gets but i need help in understanding the below

In x^7 > x^8, say we divide by x^6,
x> x^2
0>x(x-1)
Now 0>x and 0>(x-1)
for 0>x-1 ==> x <1, till here i am fine
but how do we get x>0 to make 0<x<1 ??

Till here you are fine x> x^2 so $$x^2-x<0$$ we have to solve this, and to solve let me use an old trick.
Lets solve $$x^2-x=0,x(x-1)=0$$ so x=1 or x=0. Now because the sign of x^2 is + and the operator is < we take the INTERNAL values: $$0<x<1$$.
Remember: to solve inequalities like this (x^2) treat them like equations (replace <,> with = ) then, once you have the results take a look at the sign of x^2 an the operator.
(<,-) or (>,+) take ESTERNAL values. (if they are the "same")
(>,-) or (<,+) take INTERNAL values(like this case).

Let me know if it's clear now
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10 Jun 2016, 12:53
1
AbdurRakib wrote:
Is x>0 ?

1. $$x^6$$>$$x^7$$

2. $$x^7$$ >$$x^8$$

1. $$x^6$$>$$x^7$$, either 0 < x< 1 or x < -1 (the result of even power of a negative number is positive). Not Sufficient.

2. $$x^7$$ >$$x^8$$, 0 < x< 1. x cannot be negative, If x were negative, $$x^7$$ will be negative and $$x^8$$ will be positive and this statement won't hold true. Sufficient.

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Re: Is x > 0?  [#permalink]

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14 Jun 2016, 01:06
1
Is x > 0?

(1) x^6 > x^7
(2) x^7 > x^8

$$x^6 > x^7$$

$$x^6 - x^7 > 0$$

$$x^6$$*(1 - x) > 0

$$x^6$$ is always a positive value.

so (1-x)> 0
or x < 1

x can be (1/2) a positive value or x can be -2 , a negative value, so not sufficient.

Statement (2)

$$x^7 > x^8$$

$$x^7 - x^8 > 0$$

$$x^7$$*(1 - x) > 0

$$x^6.x.(1-x)$$ > 0

Now $$x^6$$ is always a positive value.

so x(1-x)> 0

or x(x-1) < 0

0< x <1

Thus x is always positive.

B is sufficient.
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Re: Is x > 0?  [#permalink]

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17 Jun 2019, 18:31
Simply assume values. Let's say x = 0.5 or -0.5

From statement 1, X^6>X^7
Case 1, if X is 0.5, X^6 > X^7 is true
Case 2, if X is -0.5, X^>X^7 is true
We cannot prove is X is >0. Insufficient.

From Statement 2, X^7>X^8
Case 1, if X is 0.5, X^7>X^8 is true
Case 2, if X is -0.5, X^7>X^8 is false
Therefore, sufficient. X>0

Re: Is x > 0?   [#permalink] 17 Jun 2019, 18:31
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