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# Challenge - Radical Exponents (m06q20)

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Intern
Joined: 20 Apr 2011
Posts: 44
Location: United Kingdom
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10 Jun 2011, 03:19
hi, i was struggling to write ' to the power ' in symbols.
thanks
CIO
Joined: 02 Oct 2007
Posts: 1218
Followers: 94

Kudos [?]: 911 [1] , given: 334

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10 Jun 2011, 03:31
1
KUDOS
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.
thanks

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Math Forum Moderator
Joined: 20 Dec 2010
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10 Jun 2011, 03:32
1
KUDOS
arunangsude2011 wrote:
hi, i was struggling to write ' to the power ' in symbols.
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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04 Jun 2012, 21:36
(5^sqrt{2})^2 = 5^ sqrt{2} * 5^ sqrt{2}
= 5^ sqrt{2}
= 25^ sqrt{2}
Senior Manager
Joined: 15 Sep 2009
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GMAT 1: 750 Q V
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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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+1 Kudos me - I'm half Irish, half Prussian.

Intern
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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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04 Jun 2013, 05:18
The square of $$5^{\sqrt{2}}$$ = ?

A. $$5^2$$
B. $$25^{\sqrt{2}}$$
C. $$25$$
D. $$25^{2\sqrt{2}}$$
E. $$5^{\sqrt{2}^2}$$

$$(5^{\sqrt{2}})^2=5^{2*\sqrt{2}}=(5^2)^{\sqrt{2}}=25^{\sqrt{2}}$$.

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04 Jun 2013, 07:48
The answer is B. (5^sqrt{2})^2 = (5^2)^sqrt{2} = 25^sqrt{2}

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# Challenge - Radical Exponents (m06q20)

Moderator: Bunuel

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