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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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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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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.

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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Version 8.1 of the WordPress for Android app is now available, with some great enhancements to publishing: background media uploading. Adding images to a post or page? Now...

“Keep your head down, and work hard. Don’t attract any attention. You should be grateful to be here.” Why do we keep quiet? Being an immigrant is a constant...

“Keep your head down, and work hard. Don’t attract any attention. You should be grateful to be here.” Why do we keep quiet? Being an immigrant is a constant...