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# M06 P.S Q20

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Manager
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M06 P.S Q20 [#permalink]  12 Dec 2009, 11:17
In the attaced question.

Please let me know if my understanding is correct
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CIO
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Re: M06 P.S Q20 [#permalink]  17 Dec 2009, 00:02
1
KUDOS
Tania, consider these examples:

$$(5^1)^2 = 5^2 = 25$$ --> the answer is $$5^2$$, not $$25^2$$, which would equal $$5^4$$ (incorrect).

$$(5^2)^2 = 5^{2*2} = 5^4 = 625 = 25^2$$ --> You see that we had to multiply the exponents (2*2) but didn't change the base at that stage yet. If we follow your logic we end up with $$25^{2*2}$$, which is not right since we've squared the expression $$5^2$$ twice, not once (we squared the base and multiplied the exponent by 2).

Let's see our problem again:

$$(5^{\sqrt{2}})^2 = 5^{2\sqrt{2}} = (5^2)^{\sqrt{2}} = 25^{\sqrt{2}}$$ --> make sure you square the expression once

So, the right answer could be either $$25^{\sqrt{2}}$$ or $$5^{2\sqrt{2}}$$. I hope it helped make it a bit clearer.
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VP
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Re: M06 P.S Q20 [#permalink]  15 Dec 2009, 18:34
tania wrote:
In the attaced question.

Please let me know if my understanding is correct

the answer says B 25^root 2

Am I missing something or just an oversight?
CIO
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Re: M06 P.S Q20 [#permalink]  16 Dec 2009, 03:01
The correct answer is B, not D. B is exactly what you think the right answer should be .
tania wrote:
In the attaced question.

Please let me know if my understanding is correct

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Manager
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Re: M06 P.S Q20 [#permalink]  16 Dec 2009, 13:28
My bad, I mean to type the choice should have been of 25^2(root2) instead of 25^(root2) .

Can someone please explain the property of exponent for this question?
VP
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Re: M06 P.S Q20 [#permalink]  16 Dec 2009, 13:38
tania wrote:
My bad, I mean to type the choice should have been of 25^2(root2) instead of 25^(root2) .

Can someone please explain the property of exponent for this question?

Maybe it would be easier with another number

Lets say 4^sqrt 4
this would = 16

now let's say (4^sqrt4)^2 = 256

remember 5^2 * 5^3 = 5^5 because we add the exponents

so from the above example we would get 4^(2*sqrt 4)
that would in turn = 16 ^ sqrt 4 (if you wanted to solve that would equal 16*16) or 256 same as above

the point is that you add the square root not square the square root and keep the base number the same
Manager
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Re: M06 P.S Q20 [#permalink]  21 Dec 2009, 14:53
u ppl are awesome...thnx a lot for your explanations.
Intern
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Re: M06 P.S Q20 [#permalink]  15 Feb 2010, 17:46
Hey Guys, I have one quick question on this one.
By mistake i solved this one as

square of (5 ^ sqrt(2)) = (5 ^ (2 ^ (1/2) )) raised to 2.

1/2 and 2 cancels out leaving 5 ^ 2 = 25.
I am not sure whats wrong in this approach. Can someone please tell me.

Re: M06 P.S Q20   [#permalink] 15 Feb 2010, 17:46
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# M06 P.S Q20

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