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,
ds - exponents - difficult
This topic has expert replies
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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.
=> 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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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,
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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Only three values of x will satisfy the condition that x^n = x^(n+2) for any integer n: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,
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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I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.
As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.
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