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Challenge - Radical Exponents (m06q20) [#permalink ]

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12 Nov 2007, 09:11

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The square of \(5^{\sqrt{2}}\) = ?

(A) \(5^2\)

(B) \(25^{\sqrt{2}}\)

(C) \(25\)

(D) \(25^{2\sqrt{2}}\)

(E) \(5^{\sqrt{2}^2}\)

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SOLUTION: challenge-radical-exponents-m06q20-55442-20.html#p1232318

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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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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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10 Jun 2011, 04:31
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You should see this thread:

writing-mathematical-symbols-in-posts-72468.html We can display quite sophisticated mathematical expressions in our forum, like this:

\(\sqrt{(x^2 - 2x +1)^{\frac{2}{\sqrt{3}}}}\)

arunangsude2011 wrote:

hi, i was struggling to write ' to the power ' in symbols. can somebody guide me please. thanks

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

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10 Jun 2011, 04:32
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arunangsude2011 wrote:

hi, i was struggling to write ' to the power ' in symbols. can somebody guide me please. thanks

a to the power b:

a^b, ^ is called a cap and typed using "shift 6" in most keyboards.

m tag:

\(a^b\)

-1 to the power m+n

(-1)^(m+n)

m tag:

\((-1)^{(m+n)}\)

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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, 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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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^root22^2^3 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 7402 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)
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