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What is the value of x? 1) x^x = 2x 2) = x^2 is modulus

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What is the value of x? 1) x^x = 2x 2) = x^2 is modulus [#permalink] New post 23 May 2003, 18:45
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A
B
C
D
E

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What is the value of x?

1) x^x = 2x
2) [x] = x^2

[..] is modulus
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 [#permalink] New post 28 May 2003, 01:32
sorry skoper,
I think you are right......the answer should be A

I am sorry again.....don't know how i screwed up :oops: :oops: :oops:
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How ?? [#permalink] New post 28 May 2003, 12:35
How is it A ??
X^X = 2^X
equals
X^X - 2x = 0
or
X( X ^ (X-1) - 2) = 0
or
X = 0 or the solution of X^(X-1) -2 = 0 => X = 2
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 [#permalink] New post 28 May 2003, 13:14
here we go again...

if x=0 is a possible solution then you're saying x^x=0, which is generally regarded to be of indeterminate form, not 0.

what does GMAC consider?
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my view [#permalink] New post 01 Jun 2003, 21:55
This must be in the difficult pool question. Although seeming very easy to solve both statements, the question contains so many traps wait for all of those who are under pressure.


I vote this very very cool question.

Anyway, I choose A.
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Re: GMAT-DS: one more chance to catch x [#permalink] New post 15 Jul 2004, 13:20
brstorewala wrote:
What is the value of x?

1) x^x = 2x
2) [x] = x^2

[..] is modulus


Could somebody please solve this ? :thanks
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 [#permalink] New post 15 Jul 2004, 13:28
A. x^2 = 2x => x=0 or x=2 : A is insufficient

B. |x|=x^2

=> x>=0 becuase x^2 is >= 0
|x| = x if x>=0
so , equation is now x=x^2
=> x=0 or x=1

B is insufficient.

C. using both we get a unique answer x=0. C is sufficient.

HTH
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ans [#permalink] New post 15 Jul 2004, 15:24
[quote]
here we go again...

if x=0 is a possible solution then you're saying x^x=0, which is generally regarded to be of indeterminate form, not 0.

what does GMAC consider?

[/quote]

Skoper could you please explain what you mean by indeterminate form??

I think the answer is C
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 [#permalink] New post 15 Jul 2004, 15:27
for 1)
x^2=2x can be factored as x(x-2)

Therefore x=2 or x=0

Therefore 1) is insufficient.

for 2)
|x|=x^2

The only numbers that make this statement true are 0 and 1

Therefore 2) is insufficient.

So if we put the two together,
1) tells us x=2 or 0, but since 2) tells us x=1 or 0, then the combination of two results in only one answer which is 0.

I saw your posting ashkg, but needed to rephrase it this way to make sure I got the whole concept.

If I'm mistaken, somebody please let me know.
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 [#permalink] New post 15 Jul 2004, 19:55
Thats for the detailed explination Dave
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 [#permalink] New post 15 Jul 2004, 21:25
Thanks guys..

But both of you write (1) as x^2 = 2x

But infact it is x^x = 2x

Is there any specific reason for your interpretation that I am not seeing?

:thanks
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 [#permalink] New post 16 Jul 2004, 11:31
One Exerpt from the Net...

What is the value of 0^0 (^ means exponent)? Can it be considered to be the limit of a^0 as a approaches 0 ?

0^0 is indeed indeterminate. It turns out that you could make
it have any value between 0 and 1, inclusive. You could have 0 if it's the
limit as a->0 of 0^a, you could have 1 if it's the limit as a->0 of a^0,
and for x in between 0 and 1, (and this is the neat part from Dr. Shimimoto)
look at the expression (x^n)^(1/n). This just equals x for all positive
values of n. As n->Infinity, this fraction goes to 0^0, but if it's
just x the whole time, the limit of the expression as it goes to 0^0 is x.
So we could make it anything in between 0 and 1, so it's got to be
indeterminate.


Some technical crap...

In sum, I think we can't take x=0 as one of the solution.
So, the answer should be (A)..
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 [#permalink] New post 16 Jul 2004, 11:40
afife76 wrote:
Thanks guys..

But both of you write (1) as x^2 = 2x

But infact it is x^x = 2x

Is there any specific reason for your interpretation that I am not seeing?

:thanks


i screwed up.......ans should be A ( assuming 0^0 is indeterminate )
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 [#permalink] New post 16 Jul 2004, 13:02
could someone confirm that 0^0 is an unreal number as far as GMAT is concerned?

Thanks in advance
  [#permalink] 16 Jul 2004, 13:02
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