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which number is bigger (sqrt(8)^sqrt(7)) or (sqrt(7)^(sqrt(8))can we extend that for...

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fahad-alfahad | eNotes Newbie

Posted June 3, 2009 at 5:37 PM via web

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which number is bigger (sqrt(8)^sqrt(7)) or (sqrt(7)^(sqrt(8))

can we extend that for (sqrt(n+1)^sqrt(n)) or (sqrt(n)^(sqrt(n+1))

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neela | High School Teacher | Valedictorian

Posted June 4, 2009 at 4:01 AM (Answer #1)

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We know that

(n+1)^m < n^m+n^n+{nC2*n^n-2+ (n-1) lesser terms}

<2n^m+{nC2*n^(n-2)}n

<2n^m+(n^(n+1)-n^(n))/2

=n^n{(n+3)/2}<n^(n+1), for  all n>(n+3)/2 or n>3.

Thus n^n+1 > (n+1)^n=>

n^((n+1)^(1/2) > (n+1)^n^(1/2).-----------(1)

 

Using the result, n=7 and n+1=8,

7^8^(1/2) > 8^&^(1/2) is established. Take the square root and rewriting , we get,

sqrt of {7^(sqrt8)} > sqrt{8^(sqrt7)} or

{sqrt7}^(sqrt8) > {sqrt8}^(sqrt7).

 

Taking square on both sides of the relation at flag(1) we arrive at the second part which generalises the result after  changing the sides.

 

(sqrt(n+1)^sqrt(n)) < (sqrt(n)^(sqrt(n+1))

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