third root and fourth root

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third root and fourth root

by Mustang » Mon Feb 23, 2009 9:20 pm
I came across this problem in the GMATprep test

if M= Sqrt(4) + third root(4) + Fourth root (4) then M is

A. Less than 3
B. Greater then 3
C. Between 3 and 4
D. Equal to 4
E. Greater than 4

OA : E
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Re: third root and fourth root

by x2suresh » Mon Feb 23, 2009 9:24 pm
Mustang wrote:I came across this problem in the GMATprep test

if M= Sqrt(4) + third root(4) + Fourth root (4) then M is

A. Less than 3
B. Greater then 3
C. Between 3 and 4
D. Equal to 4
E. Greater than 4

OA : E
M= Sqrt(4) + third root(4) + Fourth root (4)
= 2 +(>1) +(>1)
= ( >4)

4 power of any number 0 to infinity.. : >1
4^0 =1 4^0.001 >1

so clearly 4^(1/3) > 1
4^(1/4) >1

E

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by deepoe » Thu Feb 26, 2009 5:00 am
huh?

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by BuckeyeT » Thu Feb 26, 2009 5:54 am
deepoe-

Not sure if this will help explain a bit better. But, we know that

4^(1/2) = 2
We don't know what the other values are immediately (unless you've memorized them for some reason).

BUT, we know that 1^(1/3) = 1 and 8^(1/3) = 2. So, values in between those will be greater than 1 or less than 2. So, we know that
1 < 4^(1/3) < 2

By the same logic, 1^(1/4) = 1 and 16^(1/4) = 2. So, values in between those will be greater than 1 or less than 2. So, we know that
1 < 4^(1/4) < 2

We're left with 2 + (a value between 1 and 2) + (a value between 1 and 2). Because we know the value has to be greater than 4 (as each unknown is AT LEAST greater than 1), we can confidently state E. as our answer.

Hope that helps.

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by marouan » Thu Feb 26, 2009 8:53 am
M= 4^(1/2) + 4^ (1/3) + 4 ^ (1/4)

so basicaly U have to add 1/2 + 1/3 + 1/4 = 13/12 = 1+ 1/12

this will give U:
M = 4 ^ ( 1 + 1/12)
= 4^1 + 4^ (1/12)
= 4 + 4^ (1/12)

Thus: Answer is E

Hope that helps.

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by gabriel » Thu Feb 26, 2009 9:43 am
marouan wrote:M= 4^(1/2) + 4^ (1/3) + 4 ^ (1/4)

so basicaly U have to add 1/2 + 1/3 + 1/4 = 13/12 = 1+ 1/12

this will give U:
M = 4 ^ ( 1 + 1/12)
= 4^1 + 4^ (1/12)
= 4 + 4^ (1/12)

Thus: Answer is E

Hope that helps.
I am not sure about what you have done over here. Could you please explain in a bit more detail. Why did you add the powers ??.

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by Stuart@KaplanGMAT » Thu Feb 26, 2009 11:27 am
gabriel wrote: I am not sure about what you have done over here. Could you please explain in a bit more detail. Why did you add the powers ??.
We can't add the powers in this case - we can only do so for multiplication.

2^2 + 2^3 does NOT equal 2^5
2^2 * 2^3 DOES equal 2^5

The very first explanation details one of the best way to approach this question. Let's quickly review the principles behind it:

If n is a positive number, then the nth root of ANY number greater than 1 will always be greater than 1.

This rule follows from the fact that if you multiply two positive fractions, the product is smaller than either fraction. If you multiply 1 by 1, the product stays the same. If you multiply two numbers greater than 1, you get a product greater than either number.

So, the only way to build to a bigger product is to multiply two numbers greater than 1.

For this question, we want our product to be 4.

4 can be expressed as (# bigger than 1 * # bigger than 1). Therefore, 4^(1/2) is bigger than 1.
4 can be expressed as (# bigger than 1 * # bigger than 1 * # bigger than 1). Therefore, 4^(1/3) is bigger than 1.
4 can be expressed as (# bigger than 1 * # bigger than 1 * # bigger than 1 * # bigger than 1). Therefore, 4^(1/4) is bigger than 1.

We happen to know that 4^(1/2) = 2

So, our sum will be:

2 + (# bigger than 1) + (# bigger than 1)

When we add those up, we get a number bigger than 4: choose (e).
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