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Challenge  Radical Exponents (m06q20) [#permalink]
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12 Nov 2007, 09:11
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This topic is locked. If you want to discuss this question please repost it in the respective forum. The square of \(5^{\sqrt{2}}\) = ? (A) \(5^2\) (B) \(25^{\sqrt{2}}\) (C) \(25\) (D) \(25^{2\sqrt{2}}\) (E) \(5^{\sqrt{2}^2}\) Source: GMAT Club Tests  hardest GMAT questions SOLUTION: challengeradicalexponentsm06q205544220.html#p1232318



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square of 5^(root2)=
(5^(root2)^2==
5^(root2)^2



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Re: Challenge  Radical Exponents [#permalink]
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12 Nov 2007, 11:13
bmwhype2 wrote: The square of 5^sqrt(2) is
5^2 25^sqrt(2) 25 25^2sqrt(2) 5^sqrt(2)^2
Can someone please explain the answer?
I'm getting B.
[5^sqrt(2)]^2 = (5^2)^sqrt(2) = 25^sqrt(2)



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I couldn't find the answer in the given stem.
9.75 is caculated value and none of the answer has that output
5^2 = 25
25^sqrt(2) = 94.77
25
25^2sqrt(2) = 46.08
5^sqrt(2)^2 = 25



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I get 5^2
I tried substituting 4 for 2  (5^sqrt4) x (5^sqrt4)
5^sqrt4 = 5^2 = 25  so 25 x 25 = 625 = 5^4
On that basis I'd say (5^sqrt2) x (5^sqrt2) = 5^2
Anyone with views?



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Re: Challenge  Radical Exponents [#permalink]
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13 Nov 2007, 11:43
bmwhype2 wrote: The square of 5^sqrt(2) is
5^2 25^sqrt(2) 25 25^2sqrt(2) 5^sqrt(2)^2
Can someone please explain the answer?
I get B, 25^sqrt(2)
When an exponent is raised to another exponent, they are multiplied. Thus,
The square of 5^sqrt(2) = 5^(2*sqrt(2)), then calculating backwards,
5^(2*sqrt(2)) also equals [5^(2)]^sqrt(2) = [25]^sqrt(2)



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priyankur just to let u knwo i figured out what we did wrong. we didnt square the original statement
5^(sqrt2) = 9.738 but then we have to square that to get 94.838



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Re: Challenge  Radical Exponents (m06q20) [#permalink]
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22 Dec 2009, 02:39



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Re: Challenge  Radical Exponents (m06q20) [#permalink]
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27 May 2010, 07:59
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5^\sqrt{2} x 5^\sqrt{2} => 5^\sqrt{2}+\sqrt{2} => 5^2\sqrt{2} = 25^\sqrt{2} B is correct.
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Re: Challenge  Radical Exponents (m06q20) [#permalink]
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27 May 2010, 13:54
The square of 5^sqrt(2) is
5^2 25^sqrt(2) 25 25^2sqrt(2) 5^sqrt(2)^2
Can someone please explain the answer?
B is the answer [5^sqrt(2)]^2 = 25^2sqrt(2)



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Re: Challenge  Radical Exponents (m06q20) [#permalink]
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27 May 2010, 23:49
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Here's a quote from the thread mentioned above: Quote: 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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Re: Challenge  Radical Exponents (m06q20) [#permalink]
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28 May 2010, 03:06
Answer is 25^root2 2^2^3 is same as 4^3



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Re: Challenge  Radical Exponents (m06q20) [#permalink]
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28 May 2010, 03:52
Yes, the answer is B. However, your statement in red is not correct. \(2^{2^3} = 2^8 = 256\) \(4^3 = 64\) If you meant to type \(2^{2*3}\) is same as \(4^3\), then you're correct. ameyaberi wrote: Answer is 25^root2 2^2^3 is same as 4^3
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Re: Challenge  Radical Exponents (m06q20) [#permalink]
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06 Aug 2010, 06:36
(5^sqrt2)^2 = 5^2^1/2*2^1....since same base we have to add the exponents = 5^2^3/2 = 5^sqrt2^3 = 5^2*sqrt2 = 25^sqrt2...Thus B



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Re: Challenge  Radical Exponents (m06q20) [#permalink]
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31 May 2011, 05:06
5^(root(2)) * 5^(root(2)) = 5^2(root(2)) = 25^(root(2)) Answer  B
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Re: Challenge  Radical Exponents (m06q20) [#permalink]
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31 May 2011, 08:50
answer is B
(a^m)^n = a^mn = (a^n)^m
hence
(5^sqrt2)^2 = 5^sqrt2*2 = (5^2)^sqrt2 = 25^sqrt2
hope that helps.



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Re: Challenge  Radical Exponents (m06q20) [#permalink]
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01 Jun 2011, 06:47
I get the following results.
Attachments
explantionfor5root.JPG [ 17.97 KiB  Viewed 7421 times ]



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Re: Challenge  Radical Exponents (m06q20) [#permalink]
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01 Jun 2011, 07:19
As it was correctly stated by abcg27 (you can see the quote below), you've taken the same path. If you go one step further, you'll arrive at B: \(5^{2\sqrt{2}} = (5^2)^{\sqrt{2}} = 25^{\sqrt{2}}\) abcg27 wrote: answer is B
(a^m)^n = a^mn = (a^n)^m
hence
(5^sqrt2)^2 = 5^sqrt2*2 = (5^2)^sqrt2 = 25^sqrt2
hope that helps. toughmat wrote: I get the following results.
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Re: Challenge  Radical Exponents (m06q20) [#permalink]
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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



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Re: Challenge  Radical Exponents (m06q20) [#permalink]
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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




Re: Challenge  Radical Exponents (m06q20)
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