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square root question

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by lokesh r » Wed Sep 29, 2010 1:06 pm
No.

4^(1/3)=2^(2/3)

2 x sqrt(2) = 2^(3/2)
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by amirp » Wed Sep 29, 2010 1:10 pm
thanks Lokesh,

but can't you simply 2^(2/3) one step further to get 2 sqrt 2?

I thought one of the 2's come out of the sqr and one stays inside...
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by limestone » Wed Sep 29, 2010 5:30 pm
Some rules applied to exponent:

x^a * x^b = x^(a+b)

sqrt(x) = x^(1/2) ; 3rd root (x) = x^(1/3) ; nth root (x) = x^(1/n)

(x^a)^b = x^(a*b)

Hence

4^(1/3) = (2^2)^1/3 =2^(2/3)

2 * sqrt(2) = 2^1 * 2^(1/2) = 2^(1+1/2) = 2^(3/2)

Thus the answer is NO.
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by amirp » Thu Sep 30, 2010 10:04 am
Thanks limestone,

so,,, for this question... is this the simplified version you would have?

M = 2 + 2(2/3) + 2 (1/2)

or can it be simplified one more step?
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by GMATGuruNY » Thu Sep 30, 2010 11:01 am
amirp wrote:Thanks limestone,

so,,, for this question... is this the simplified version you would have?

M = 2 + 2(2/3) + 2 (1/2)

or can it be simplified one more step?
Since the problem above asks only for an estimation, we can just ballpark:

√4 = 2.
4^(1/3) > 1.
4*(1/4) = √2 ≈ 1.4.

So M > 4.4.
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by ankur.agrawal » Thu Sep 30, 2010 11:23 am
GMATGuruNY wrote:
amirp wrote:Thanks limestone,

so,,, for this question... is this the simplified version you would have?

M = 2 + 2(2/3) + 2 (1/2)

or can it be simplified one more step?
Since the problem above asks only for an estimation, we can just ballpark:

√4 = 2.
4^(1/3) > 1.
4*(1/4) = √2 ≈ 1.4.

So M > 4.4.
Can we do it by taking common:

4^1/2[ 1+4^(2/3) +4^(1/2)]---- solving this we can clearly see that it is greater than 4
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by GMATGuruNY » Thu Sep 30, 2010 11:31 am
ankur.agrawal wrote:
GMATGuruNY wrote:
amirp wrote:Thanks limestone,

so,,, for this question... is this the simplified version you would have?

M = 2 + 2(2/3) + 2 (1/2)

or can it be simplified one more step?
Since the problem above asks only for an estimation, we can just ballpark:

√4 = 2.
4^(1/3) > 1.
4*(1/4) = √2 ≈ 1.4.

So M > 4.4.
Can we do it by taking common:

4^1/2[ 1+4^(2/3) +4^(1/2)]---- solving this we can clearly see that it is greater than 4
Sure, but I think that factoring out 4^1/2 is time-consuming and makes the problem more complicated. Estimating is quicker and safer.
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by amirp » Thu Sep 30, 2010 12:36 pm
Thanks GmatGuru,

I see how your approach makes total sense. for the sake of learning the concept tho, can you explain how far I can simplify

4^(1/3)?

I know I can get 2^(2/3) but, for some reason I keep thinking I can simplify it even more. can you explain?
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by GMATGuruNY » Thu Sep 30, 2010 1:11 pm
amirp wrote:Thanks GmatGuru,

I see how your approach makes total sense. for the sake of learning the concept tho, can you explain how far I can simplify

4^(1/3)?

I know I can get 2^(2/3) but, for some reason I keep thinking I can simplify it even more. can you explain?
There are some problems in which you might want to raise each side of the equation to the reciprocal power:

x^(2/3) = 4
(x^(2/3))^3/2 = 4^(3/2)
x^(6/6) = 8
x = 8

Is this what you mean?
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by amirp » Thu Sep 30, 2010 1:23 pm
No.

I just keep thinking that since I have a 2^2 inside the cube root, I can take it out and solve it one step further...

4^(1/3)=
2^(2/3)=
( 2^(1/3) )^2 =
and more simplification here? (example: 2 sqr 2 )

Maybe i'm over thinking it and the simplest form is 2^(2/3) ??
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