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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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It is beyond a doubt that all our knowledge that begins with experience.

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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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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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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Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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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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