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Re: If x is positive and not equal to 1, then the product of x^(1/n) for a [#permalink]
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Bunuel wrote:
If x is positive and not equal to 1, then the product of x^(1/n) for all positive integers n such that 21 ≤ n ≤ 30 is between

A. 1 and x^(1/6)
B. x^(1/6) and x^(1/3)
C. x^(1/3) and x^(1/2)
D. x^(1/2) and x^(2/3)
E. x^(2/3) and x^(5/6)


Kudos for a correct solution.


x^(1/21) + x^(1/22) + ... + x^(1/30)

S = 1/21 + 1/22 + ... + 1/30

or S > 1/21 + 1/21 ... + 1/21
S > 10/21
and S < 10/30

10/21 < S < 1/3

so C
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Re: If x is positive and not equal to 1, then the product of x^(1/n) for a [#permalink]
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x is positive and not equal to 1
Product of x^(1/n) for all positive integers n such that 21 ≤ n ≤ 30 is

= \(x^{(\frac{1}{21}+\frac{1}{29})+(\frac{1}{22}+\frac{1}{28})+(\frac{1}{23}+\frac{1}{27})+(\frac{1}{24}+\frac{1}{26})+\frac{1}{25}+\frac{1}{30}}\)

= \(x^{(\frac{1}{20+1}+\frac{1}{20+9})+(\frac{1}{20+2}+\frac{1}{20+8})+(\frac{1}{20+3}+\frac{1}{20+7})+(\frac{1}{20+4}+\frac{1}{20+6})+\frac{1}{25}+\frac{1}{30}}\)

= \(x^{\frac{1}{12}+\frac{1}{12}+\frac{1}{12}+\frac{1}{12}+\frac{1}{25}+\frac{1}{30}}\)

= \(x^{\frac{61}{150}}\)

= \(x^{\frac{2}{5}}\)

\(x^{\frac{1}{3}} < x^{\frac{2}{5}} < x^{\frac{1}{2}}\)

Answer C
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Re: If x is positive and not equal to 1, then the product of x^(1/n) for a [#permalink]
thefibonacci wrote:

x^(1/21) + x^(1/22) + ... + x^(1/30)

S = 1/21 + 1/22 + ... + 1/30

or S > 1/21 + 1/21 ... + 1/21
S > 10/21
and S < 10/30

10/21 < S < 1/3

so C


Should not it be the other way around? 1/3 < S < 1/2
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Re: If x is positive and not equal to 1, then the product of x^(1/n) for a [#permalink]
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Re: If x is positive and not equal to 1, then the product of x^(1/n) for a [#permalink]
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