Is x > 0?

[BTGmoderatorDC]

Is x > 0?

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

OA B

Source: Veritas Prep

[Jay@ManhattanReview]

Is x > 0?

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

OA B

Source: Veritas Prep

Let's take each statement one by one.

(1) x^6 > x^7

=> 1 > x; cancelling x^6 from both the sides. Since the exponent of x^6 is 6, an even number, the value of x^6 must be positive, irrespective of x is negative or positive, making us need not worry about the change in sign of the inequality.

So, we have 1 > x,

If 1 > x > 0, the answer is Yes; however if x < 0, the answer is No. Insufficient.

(2) x^7 > x^8

=> x > x^2; cancelling x^6 from both the sides. We cannot divide both the sides by x^7 since the exponent of x^7 is 7, an odd number. Whether x^7 is positive or negative will depend on the value of x. If x is positive, x^7 is positive; however, if x is negative, x^7 is negative, making us reverse the sign of inequality.

So, we have x > x^2.

Note that x cannot be negative since x^2 is a positive term, it cannot be less than a negative term, x. Thus, x is a positive term. Or, x > 0. Sufficient.

The correct answer: B

Hope this helps!

-Jay
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