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# 2^199 + 2^199 = 2^X

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Manager
Joined: 10 Jun 2003
Posts: 210
Location: Maryland
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Kudos [?]: 5 [0], given: 0

2^199 + 2^199 = 2^X [#permalink]  29 Aug 2003, 06:30
00:00

Difficulty:

(N/A)

Question Stats:

0% (00:00) correct 0% (00:00) wrong based on 0 sessions
Solve for X

2^199 + 2^199 = 2^X
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Sept 3rd

Manager
Joined: 10 Jun 2003
Posts: 210
Location: Maryland
Followers: 2

Kudos [?]: 5 [0], given: 0

yes, that is great!!

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Sept 3rd

Manager
Joined: 10 Jun 2003
Posts: 210
Location: Maryland
Followers: 2

Kudos [?]: 5 [0], given: 0

Surely, that must have been confusing. Here we go...

3^199 + 3^199 = 3^X
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Sept 3rd

Intern
Joined: 17 Aug 2003
Posts: 38
Location: USA
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Kudos [?]: 1 [0], given: 0

mciatto wrote:
Surely, that must have been confusing. Here we go...

3^199 + 3^199 = 3^X

I don't even know if you can solve 3^199 + 3^199 = 2^X without a calculator i just didn't want to assume.

Anyway i'm really stumped on this one. The way i did the previous one was:-

2^199 + 2^199 = 2^X
2(2^199) = 2^X
2^199 = 2^X-1

Therefore X has to be 200. However this method does not work for the 3^199 question, unless i'm missing something, which would not be the first time.
Manager
Joined: 14 Aug 2003
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Location: barcelona
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3^199 + 3^199 = 3^X

im not sure if we can solve that without a calculator... id do the following:

3^199 + 3^199=2*(3^199)=3^X

2=3^y, solving for y we get y=log2/log3

thus 2*(3^199)=3^(log2/log3)*(3^199)=3^(log2/log3+199)

x=log2/log3+199

i may be mistaken though...
Manager
Joined: 10 Jun 2003
Posts: 210
Location: Maryland
Followers: 2

Kudos [?]: 5 [0], given: 0

I think this is obviously too complicated for the GMAT, but it was worth a shot.

All I can say is that

3^199 + 3^199 = (2/3)* 3^200
I'm not sure if this equates to javropu's solution.
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Sept 3rd

Manager
Joined: 10 Jun 2003
Posts: 210
Location: Maryland
Followers: 2

Kudos [?]: 5 [0], given: 0

I guess all that's for sure is 199<X<200.

Hopefully there would be one answer choice in that range
_________________

Sept 3rd

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