No.
4^(1/3)=2^(2/3)
2 x sqrt(2) = 2^(3/2)
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square root question
Source: Beat The GMAT — Problem Solving |
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.
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.
"There is nothing either good or bad - but thinking makes it so" - Shakespeare.
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GMATGuruNY
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Since the problem above asks only for an estimation, we can just ballpark: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?
√4 = 2.
4^(1/3) > 1.
4*(1/4) = √2 ≈ 1.4.
So M > 4.4.
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ankur.agrawal
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Can we do it by taking common:GMATGuruNY wrote:Since the problem above asks only for an estimation, we can just ballpark: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?
√4 = 2.
4^(1/3) > 1.
4*(1/4) = √2 ≈ 1.4.
So M > 4.4.
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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GMATGuruNY
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Sure, but I think that factoring out 4^1/2 is time-consuming and makes the problem more complicated. Estimating is quicker and safer.ankur.agrawal wrote:Can we do it by taking common:GMATGuruNY wrote:Since the problem above asks only for an estimation, we can just ballpark: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?
√4 = 2.
4^(1/3) > 1.
4*(1/4) = √2 ≈ 1.4.
So M > 4.4.
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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I unlock the best way for YOU to solve problems.
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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?
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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GMATGuruNY
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There are some problems in which you might want to raise each side of the equation to the reciprocal power: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?
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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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) ??
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) ??
