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Challenge - Radical Exponents (m06q20)

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Re: Challenge - Radical Exponents (m06q20) [#permalink] New post 10 Jun 2011, 03:19
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] 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] 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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Re: Challenge - Radical Exponents (m06q20) [#permalink] New post 04 Jun 2012, 21:36
Answer : 25^sqrt{2}
(5^sqrt{2})^2 = 5^ sqrt{2} * 5^ sqrt{2}
= 5^ sqrt{2}
= 25^ sqrt{2}
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Re: Challenge - Radical Exponents (m06q20) [#permalink] New post 05 Jun 2012, 03:16
B is the answer. Tricky one but remember that the exponents will multiply each other, giving you 5^(2root2).

Cheers.

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Re: Challenge - Radical Exponents (m06q20) [#permalink] New post 13 Sep 2012, 07:35
a^b^c is a^(b^c) and not a^bc.So ,root(2)^2=1.9946~2;5^2=25
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Re: Challenge - Radical Exponents (m06q20) [#permalink] New post 04 Jun 2013, 05:18
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Re: Challenge - Radical Exponents (m06q20) [#permalink] New post 04 Jun 2013, 07:48
The answer is B. (5^sqrt{2})^2 = (5^2)^sqrt{2} = 25^sqrt{2} :)
Re: Challenge - Radical Exponents (m06q20)   [#permalink] 04 Jun 2013, 07:48
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