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# m06 Q20

Author Message
SVP
Joined: 16 Nov 2010
Posts: 1597

Kudos [?]: 592 [0], given: 36

Location: United States (IN)
Concentration: Strategy, Technology

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12 Apr 2011, 02:56
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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Kudos [?]: 592 [0], given: 36

Math Forum Moderator
Joined: 20 Dec 2010
Posts: 1964

Kudos [?]: 2051 [1], given: 376

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12 Apr 2011, 03:09
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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Kudos [?]: 2051 [1], given: 376

Senior Manager
Joined: 08 Nov 2010
Posts: 394

Kudos [?]: 128 [0], given: 161

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13 Apr 2011, 10:48
yea, i would be happy to find that kind of practice if someone have it.

thanks.
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Kudos [?]: 128 [0], given: 161

Director
Joined: 01 Feb 2011
Posts: 726

Kudos [?]: 143 [0], given: 42

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

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

Posted from my mobile device

Kudos [?]: 143 [0], given: 42

Re: m06 Q20   [#permalink] 15 Apr 2011, 15:47
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# m06 Q20

Moderator: Bunuel

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