Which of the following is a value of x for which : Problem Solving (PS)

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MasteringGMAT

Which of the following is a value of x for which $x^{-9} –x^{-7} > 0$?

A. -2
B. -1
C. $-\frac{1}{2}$
D. 1
E. 2

Using the exponent rules, we have:

1/x^9 – 1/x^7 > 0

We see that x cannot be equal to zero.

If x > 0, then:

1 - x^9/x^7 > 0

1 - x^2 > 0

1 > x^2

|x| < 1 => -1 < x < 1

So, in this case, it must be true that 0 < x < 1.

If x < 0, then:

1 - x^9/x^7 < 0

|x| > 1 => x > 1 OR x < -1

So, in this case, it must be true that x < -1.

Therefore, x < -1 OR 0 < x < 1.

Answer: A

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Updated on: 18 Jan 2024, 11:57

Originally posted by flyinghirsch on 29 Dec 2023, 13:33.
Last edited by flyinghirsch on 18 Jan 2024, 11:57, edited 1 time in total.

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--- SIMPLE SOLUTION ---

Even though it does not look like it, the question is basically about neg * pos * neg:

Step 1: Factor out X^-9
X^-9 * (1 - x^2) > 0

Step 2: Difference of Square formula
X^-9*(1 - x)*(1+x) > 0

Step 3: Test cases
Essentially product of 3 factors must be pos:
For B) and D) solution of 3 factors would be 0 -> eliminate
For A) equation become (1/-2^9) * ( 1--2) * (1-2) -> sufficient as neg * pos * neg is positive and therefore > 0

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09 Jan 2024, 00:17

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x^-9 - x^-7 > 0
x^-9 * (1 - x^2) > 0
x^-9 * (1 - x) * (1 + x) > 0
We get the roots as 0, 1, or -1.

The critical points 0, 1, and -1 divide the number line into four intervals as shown. Let’s select a test point from each interval to determine the sign of the inequality within that interval, then we will obtain the answer.

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gmatophobia

MasteringGMAT

Which of the following is a value of x for which $x^{-9} –x^{-7}$ > 0?

A. -2
B. -1
C. -$\frac{1}{2}$
D. 1
E. 2

$x^{-9} –x^{-7}$ > 0

$x^{-9}(1 –x^{2})$ > 0

Case 1: $x > 0$

If $x > 0 → 1 - x^2 > 0$

$ 1 - x^2 > 0$

$ x^2 < 1$

We can eliminate options D, and E as in both the options $x > 0 $ and $ x^2 \geq 1$

Case 2: $x < 0$

If $x < 0 → 1 - x^2 < 0$

$ 1 - x^2 < 0$

$ x^2 > 1$

Options A, B, and C have values of $x < 0$. However, among the three options, only in option A the value of $x^2 > 1$.

Hence, Option A is correct.

Please help. What is wrong with the way I am doing it? I am stuck after the case II in the paper solution below. How do I interpret this solution in terms of answer choices? Any feedback will help. Thank you. chetan2u KarishmaB

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chetan2u

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26 Jan 2024, 09:48

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Engineer1

Please help. What is wrong with the way I am doing it? I am stuck after the case II in the paper solution below. How do I interpret this solution in terms of answer choices? Any feedback will help. Thank you. chetan2u KarishmaB

$x^{-7}>0……\frac{1}{x^7}>0…..x^7>0$

You are going wrong in changing the inequality sign.
You do not change the sign of power and inequality sign. The negative powers just take the reciprocal but have no contribution in change of sign except that even powers are always positive.

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24 Feb 2024, 02:41

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Can someone or @JeffTargetTestPrep explain how you went from 1/x^9 - 1/x^7 > 0 to 1 - x^2 > 0 ? I'm lost here.

JeffTargetTestPrep

MasteringGMAT

Which of the following is a value of x for which $x^{-9} –x^{-7} > 0$?

A. -2
B. -1
C. $-\frac{1}{2}$
D. 1
E. 2

Using the exponent rules, we have:

1/x^9 – 1/x^7 > 0

We see that x cannot be equal to zero.

If x > 0, then:

1 - x^9/x^7 > 0

1 - x^2 > 0

1 > x^2

|x| < 1 => -1 < x < 1

So, in this case, it must be true that 0 < x < 1.

If x < 0, then:

1 - x^9/x^7 < 0

|x| > 1 => x > 1 OR x < -1

So, in this case, it must be true that x < -1.

Therefore, x < -1 OR 0 < x < 1.

Answer: A

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kevinhirose

Can someone or @JeffTargetTestPrep explain how you went from 1/x^9 - 1/x^7 > 0 to 1 - x^2 > 0 ? I'm lost here.

JeffTargetTestPrep

MasteringGMAT

Which of the following is a value of x for which $x^{-9} –x^{-7} > 0$?

A. -2
B. -1
C. $-\frac{1}{2}$
D. 1
E. 2

Using the exponent rules, we have:

1/x^9 – 1/x^7 > 0

We see that x cannot be equal to zero.

If x > 0, then:

1 - x^9/x^7 > 0

1 - x^2 > 0

1 > x^2

|x| < 1 => -1 < x < 1

So, in this case, it must be true that 0 < x < 1.

If x < 0, then:

1 - x^9/x^7 < 0

|x| > 1 => x > 1 OR x < -1

So, in this case, it must be true that x < -1.

Therefore, x < -1 OR 0 < x < 1.

Answer: A

X^-9 = 1/X^9 and X^-7=1/X^7 so
the problem can be restated as:

1/X^9 - 1/X^7 > 0

Multiply both sides by X^9:

X^9(1/X^9 - 1/X^7) > 0×X^9

= 1 - X9/X^7 > 0 =

1-X^2 > 0

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21 Nov 2024, 01:42

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if we substitute the answers why cant we arrive at the answer ?

the value of x satisfies both -2 & -1/2??
which is the correct answer??
KarishmaB GMATNinja bb Bunuel

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naveeng15

if we substitute the answers why cant we arrive at the answer ?

the value of x satisfies both -2 & -1/2??
which is the correct answer??
KarishmaB GMATNinja bb Bunuel

The correct answer is given under the spoiler in the original post and it's A, -2.

-1/2 does not satisfy $x^{(-9)} - x^{(-7)} > 0$:

$(-\frac{1}{2})^{(-9)} - (-\frac{1}{2})^{(-7)} =$

$=(-2)^9 - (-2)^7 =$

$=-512 + 128 =$

$=-384$

P.S. As explained above $x^{(-9)} - x^{(-7)} > 0$ holds for x < -1 and 0 < x < 1.

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look - thumb rule is that 2^9 is > 2^7, but in case of decimal, 1/2^9 < 1/2^7. Now as for the 1st option i.e. -2, substitution gives us -(1/2^9) + 1/2^7 i.e, 1/2^7 - 1/2^9, since we know that 1/2^7 > 1/2^9, the answer will be >0

kevinhirose

Can someone or @JeffTargetTestPrep explain how you went from 1/x^9 - 1/x^7 > 0 to 1 - x^2 > 0 ? I'm lost here.

JeffTargetTestPrep

MasteringGMAT

Which of the following is a value of x for which $x^{-9} –x^{-7} > 0$?

A. -2
B. -1
C. $-\frac{1}{2}$
D. 1
E. 2

Using the exponent rules, we have:

1/x^9 – 1/x^7 > 0

We see that x cannot be equal to zero.

If x > 0, then:

1 - x^9/x^7 > 0

1 - x^2 > 0

1 > x^2

|x| < 1 => -1 < x < 1

So, in this case, it must be true that 0 < x < 1.

If x < 0, then:

1 - x^9/x^7 < 0

|x| > 1 => x > 1 OR x < -1

So, in this case, it must be true that x < -1.

Therefore, x < -1 OR 0 < x < 1.

Answer: A

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07 Jan 2025, 00:09

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One way to solve this.

Harsha

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MasteringGMAT

Which of the following is a value of x for which $x^{-9} –x^{-7} > 0$?

A. -2
B. -1
C. $-\frac{1}{2}$
D. 1
E. 2

$x^{-9} –x^{-7} > 0$

Take x^-9 common

$x^{-9} (1 - x^2) > 0$

$\frac{(1-x^2)}{x^9} > 0\\
$

Now, x cannot be 1 or -1 because it makes the LHS 0.

If x = -2, both numerator and denominator of LHS are negative so the fraction becomes positive. Hence this is correct.

Note that for both 2 and - 1/2, the LHS will be negative.

Answer (A)

Method 2:

If you want to solve it using the wavy line method,
$\frac{(x^2-1)}{x^9} < 0$

$\frac{(x +1)(x-1)}{x^9} < 0$
Transition points are -1, 0 and 1 so the expression will be negative when x is between 0 and 1 OR less than -1.

Only option (A) works.

Wavy line method is discussed here: https://youtu.be/PWsUOe77__E

Which of the following is a value of x for which

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