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# Challenge - Radical Exponents (m06q20)

Author Message
Intern
Joined: 24 May 2010
Posts: 46

Kudos [?]: 5 [0], given: 6

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01 Jun 2011, 07:25
toughmat wrote:
I get the following results.

^^ what you did so far is correct. Further to match with the given choices, split $$5^2*\sqrt{^2}$$ as $$(5^2)\sqrt{^2} = (25)\sqrt{^2}$$, which is given in choices

Kudos [?]: 5 [0], given: 6

Intern
Joined: 20 Apr 2011
Posts: 40

Kudos [?]: [0], given: 9

Location: United Kingdom

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10 Jun 2011, 04:18
(a^x)^m = (a^m)^x
(5^[square_root]2)^2 = (5^2)^[square_root]2 = 25^[square_root]2

Kudos [?]: [0], given: 9

Intern
Joined: 20 Apr 2011
Posts: 40

Kudos [?]: [0], given: 9

Location: United Kingdom

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10 Jun 2011, 04:19
hi, i was struggling to write ' to the power ' in symbols.
thanks

Kudos [?]: [0], given: 9

Intern
Joined: 20 Apr 2012
Posts: 1

Kudos [?]: 5 [0], given: 16

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04 Jun 2012, 22:36
(5^sqrt{2})^2 = 5^ sqrt{2} * 5^ sqrt{2}
= 5^ sqrt{2}
= 25^ sqrt{2}

Kudos [?]: 5 [0], given: 16

Senior Manager
Joined: 15 Sep 2009
Posts: 263

Kudos [?]: 76 [0], given: 6

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05 Jun 2012, 04:16
B is the answer. Tricky one but remember that the exponents will multiply each other, giving you 5^(2root2).

Cheers.
_________________

+1 Kudos me - I'm half Irish, half Prussian.

Kudos [?]: 76 [0], given: 6

Intern
Joined: 19 Aug 2012
Posts: 3

Kudos [?]: [0], given: 3

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13 Sep 2012, 08:35
a^b^c is a^(b^c) and not a^bc.So ,root(2)^2=1.9946~2;5^2=25

Kudos [?]: [0], given: 3

Math Expert
Joined: 02 Sep 2009
Posts: 41875

Kudos [?]: 128470 [0], given: 12177

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04 Jun 2013, 06:18
The square of $$5^{\sqrt{2}}$$ = ?

A. $$5^2$$
B. $$25^{\sqrt{2}}$$
C. $$25$$
D. $$25^{2\sqrt{2}}$$
E. $$5^{\sqrt{2}^2}$$

$$(5^{\sqrt{2}})^2=5^{2*\sqrt{2}}=(5^2)^{\sqrt{2}}=25^{\sqrt{2}}$$.

_________________

Kudos [?]: 128470 [0], given: 12177

Intern
Joined: 30 May 2013
Posts: 13

Kudos [?]: 6 [0], given: 2

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04 Jun 2013, 08:48
The answer is B. (5^sqrt{2})^2 = (5^2)^sqrt{2} = 25^sqrt{2}

Kudos [?]: 6 [0], given: 2

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# Challenge - Radical Exponents (m06q20)

Moderator: Bunuel

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