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# m06 Q20

Author Message
SVP
Joined: 16 Nov 2010
Posts: 1672
Location: United States (IN)
Concentration: Strategy, Technology
Followers: 33

Kudos [?]: 514 [0], given: 36

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12 Apr 2011, 01:56
Hi

For this question, I calculated as follows :

5^(root(2)) * 5^(root(2))

= 5^(2root(2))

= 25(root(2))

Is the above approach correct ? Also, where can I learn more about such tricky exponent stuff ?

Regards,
Subhash
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Joined: 20 Dec 2010
Posts: 2021
Followers: 161

Kudos [?]: 1703 [1] , given: 376

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12 Apr 2011, 02:09
1
KUDOS
subhashghosh wrote:
Hi

For this question, I calculated as follows :

5^(root(2)) * 5^(root(2))

= 5^(2root(2))

= 25(root(2))

Is the above approach correct ? Also, where can I learn more about such tricky exponent stuff ?

Regards,
Subhash

Yes, the approach is correct.

$$(5^{\sqrt{2}})^2=(5^{2})^{\sqrt{2}}=25^{\sqrt{2}}$$ [Note: $$(x^m)^n=x^{mn}=(x^n)^m$$]

I think GMAT Club Math book exponents in number theory did a decent job.
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Senior Manager
Joined: 08 Nov 2010
Posts: 417
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Kudos [?]: 105 [0], given: 161

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13 Apr 2011, 09:48
yea, i would be happy to find that kind of practice if someone have it.

thanks.
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Director
Joined: 01 Feb 2011
Posts: 757
Followers: 14

Kudos [?]: 115 [0], given: 42

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15 Apr 2011, 14:47
Looks alright to me.

I got 25^(sqrt(2)) too.

Posted from my mobile device
Re: m06 Q20   [#permalink] 15 Apr 2011, 14:47
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# m06 Q20

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