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The square of 5^sqrt(2) = ? a) 5^2 b) 25^sqrt(2) c) 25

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Director
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The square of 5^sqrt(2) = ? a) 5^2 b) 25^sqrt(2) c) 25  [#permalink]  17 Jul 2007, 14:37
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The square of 5^sqrt(2) = ?

a) 5^2
b) 25^sqrt(2)
c) 25
d) 25^2sqrt(2)
e) 5^sqrt(2)^2

Director
Joined: 26 Feb 2006
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Re: PS: Square of 5^sqrt(2) [#permalink]  17 Jul 2007, 14:51
GK_Gmat wrote:
The square of 5^sqrt(2) = ?

a) 5^2
b) 25^sqrt(2)
c) 25
d) 25^2sqrt(2)
e) 5^sqrt(2)^2

i suppose sqrt (5) [or 5^(1/2)] and 5^sqrt(2) are different from each other.

= [5^sqrt(2)}^2
= 5^sqrt(2) x 5^sqrt(2)
= 5^2(sqrt(2))
= 5^sqrt(2)^2
Manager
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Re: PS: Square of 5^sqrt(2) [#permalink]  17 Jul 2007, 18:44
Himalayan wrote:
GK_Gmat wrote:
The square of 5^sqrt(2) = ?

a) 5^2
b) 25^sqrt(2)
c) 25
d) 25^2sqrt(2)
e) 5^sqrt(2)^2

i suppose sqrt (5) [or 5^(1/2)] and 5^sqrt(2) are different from each other.

= [5^sqrt(2)}^2
= 5^sqrt(2) x 5^sqrt(2)
= 5^2(sqrt(2))
= 5^sqrt(2)^2

Himalayan, can you please explain how did you get
5^2(sqrt(2)) = 5^sqrt(2)^2.

Basically, I am struglling with how is A^xy = A^y^x. What am I missing?
Intern
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(5^sqrt(2))^2=5^(2sqrt(2))=25^sqrt(2)

(B)
VP
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n^x^y = (n^x)^y and not n^(x^y)

5^sqrt(2)^2 =

(5^sqrt(2))^2 =

5^sqrt(2)*5^sqrt(2) =

25^sqrt(2) =

the same way that

2^4^2 = (2^4)^2 = (2^4)*(2^4) = 4^4

Senior Manager
Joined: 14 Jun 2007
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KillerSquirrel wrote:
n^x^y = (n^x)^y and not n^(x^y)

5^sqrt(2)^2 =

(5^sqrt(2))^2 =

5^sqrt(2)*5^sqrt(2) =

25^sqrt(2) =

the same way that

2^4^2 = (2^4)^2 = (2^4)*(2^4) = 4^4

so the rule is x^a*x^b = x^a+b
therefore 5^2sqrt(2)... so we can take the first two and square the five and leave sqrt2 still up there? that is interesting... i never knew that! thanks squirrel.

as for the other rule, it is x^a^b =(x^a)^b=x^ab
VP
Joined: 08 Jun 2005
Posts: 1146
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Kudos [?]: 136 [0], given: 0

Anonymousegmat you are correct:

x^n^y = x^(n*y)

3^2^4 = 3^8 = 6,561

Nice to remember !

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