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Challenge - Radical Exponents (m06q20)

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Challenge - Radical Exponents (m06q20) [#permalink]

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New post 12 Nov 2007, 08:11
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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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New post 27 May 2010, 22:49
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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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New post 27 May 2010, 06:59
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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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New post 10 Jun 2011, 03:31
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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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New post 10 Jun 2011, 03:32
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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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New post 12 Nov 2007, 08:49
square of 5^(root2)=

(5^(root2)^2==

5^(root2)^2

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Re: Challenge - Radical Exponents [#permalink]

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New post 12 Nov 2007, 10:13
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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New post 12 Nov 2007, 10:39
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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New post 13 Nov 2007, 09:50
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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New post 13 Nov 2007, 10:43
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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New post 13 Nov 2007, 10:50
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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New post 22 Dec 2009, 01:39
Here's another discussion related to this question:

m06-p-s-q20-87912.html
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Re: Challenge - Radical Exponents (m06q20) [#permalink]

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New post 27 May 2010, 12:54
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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New post 28 May 2010, 02:06
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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New post 28 May 2010, 02:52
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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New post 06 Aug 2010, 05:36
(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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New post 31 May 2011, 04:06
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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New post 31 May 2011, 07:50
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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New post 01 Jun 2011, 05:47
I get the following results.
Attachments

explantionfor5root.JPG
explantionfor5root.JPG [ 17.97 KiB | Viewed 7483 times ]

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

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New post 01 Jun 2011, 06:19
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] 01 Jun 2011, 06:19

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