ds - exponents - difficult

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ds - exponents - difficult

by ccassel » Wed Apr 06, 2011 10:41 am
Hi,

How would you answer this question?

x^n = x^(n+2) for any integer n. Is it true that x>0?

(1) x = x^2 -2
(2) 2x < x^5

Cheers,

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by maihuna » Wed Apr 06, 2011 11:24 am
x^n = x^(n+2)
=> X^n(x^2-1) = 0
either x=0, or x^2=1=> x=+1, -1 so x=0, 1 or -1

1. x^2-x = 2 => x(x-1) = 2 => x=2, or x = -1
combining know values, x = -1, so suff, that x>0 answered in neg

2. 2x < x^5 => x(2-x^4)<0, x<0 and 2>x^4 =>x^4<2 => x<+/-2^1/4 since x<0, x<-2^1/4
or x>0 and 2<x^4 or x^4 > 2 so x>+/-2^1/4, since x>0, x>2^1/4

This doesn't answer as none of the values are consistent.
Last edited by maihuna on Wed Apr 06, 2011 1:23 pm, edited 1 time in total.
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by ccassel » Wed Apr 06, 2011 12:06 pm
Statement 1 looks correct to me as well.

However, can you explain Statement 2? Why did you replace ">" with "=" sign? Also, why is -2^1/4 & x>0 sufficient to answer Statement 2 as correct? Would'nt that give solutions to both sides of the question making Statement 2 not sufficient?

"2x < x^5 => x(x^4-2)=0, x=0, or x=+/-2^1/4"

This is how I solved the statement 2...
2x<x^5
0<x^5-2x
x(x^4-2)>0
therefore,

x>0

and

x^4-2>0
x^4>2
x>+/- 1/2^4
x>+/- 1/16

x>1/16
and
x>-1/16

Therefore, statement 2 is not sufficient to answer is x>0.

What am I missing? Is there an exception to be made with quadratics and an inequality?

Thanks,

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by maihuna » Wed Apr 06, 2011 1:24 pm
I didn't noticed the <, I have edited my response, if it is wrong do repost the OA and Q again, may be there is some typo.
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by ccassel » Wed Apr 06, 2011 1:37 pm
The answer is D if anyone can explain how you got there.

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by GMATGuruNY » Wed Apr 06, 2011 6:34 pm
ccassel wrote:Hi,

How would you answer this question?

x^n = x^(n+2) for any integer n. Is it true that x>0?

(1) x = x^2 -2
(2) 2x < x^5

Cheers,
Only three values of x will satisfy the condition that x^n = x^(n+2) for any integer n:
x=-1, x=0, x=1.

Let's plug x=-1, x=0, and x=1 into x^n = x^(n+2).

If n=2 and n+2=4:
(-1)² = (-1)�. Yes.
0² = 0�. Yes.
1² = 1�. Yes.

If n=3 and n+2=5:
(-1)³ = (-1)�. Yes.
0³ = 0�. Yes.
1³ = 1�. Yes.

Thus, our only options are x=-1, x=0, or x=1.
Since we want to know whether x>0, the question can be rephrased: Does x=1?

Statement 1: x = x² - 2.
x² - x - 2 = 0.
(x-2)(x+1) = 0.
x=2 or x=-1.
Thus, it is not true that x=1.
Sufficient.

Statement 2: 2x < x�.
Of the 3 possible values x=-1, x=0, and x=1, only x=-1 satisfies statement 2:
2(-1) < (-1)�
-2 < -1.
Thus, it is not true that x=1.
Sufficient.

The correct answer is D.
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