Express (log n)^(log n) as a power of n,

[Jamin2112]

Homework Statement

Express (log n)^(log n) as a power of n, and use the result to show ∑1/(log n)^(log n) is convergent.

Homework Equations

?

The Attempt at a Solution

I've heard of an alternative way to express ln(n), so I cannot get started.

 

[Deadstar]

Way I just did it was...

n^{f(n)} = \log(n)^{\log(n)}

take log of both sidesf(n) \log(n) = \log(n) \log(\log(n))

And f(n) is pretty obvious from this line.

 

[Jamin2112]

Way I just did it was...

n^{f(n)} = \log(n)^{\log(n)}

take log of both sides


f(n) \log(n) = \log(n) \log(\log(n))

And f(n) is pretty obvious from this line.

f(n) = n ?

 

[Petek]

What do you get if you divide both sides of Deadstar's last equation by log(n)?

 

[Jamin2112]

What do you get if you divide both sides of Deadstar's last equation by log(n)?

You get f(n) = log(log n), which is equal to n.

 

[michael.wes]

You get f(n) = log(log n), which is equal to n.

That isn't right.. check your work

 

[micromass]

Why would n=log(log(n))?

 

[Petek]

You get f(n) = log(log n), which is equal to n.

Perhaps you're thinking of exp(log(n)) or log(exp(n))?

 

[Dick]

You get f(n) = log(log n), which is equal to n.

f(n)=log(log(n)) is correct. log(log(n))=n is complete baloney. I don't know why you'd even say that. If you'd been paying any attention to Deadstar you'd now have log(n)^log(n)=n^(log(log(n))). That really ought to tell you something pretty quickly about the convergence question.

 

[gomunkul51]

ln(ln(X)) IS NOT X.

you were thinking of:
ln(exp(x)) = exp(ln(x)) = x

 

[arildno]

We have:
\log(n)^{\log(n)}=(e^{\log(\log(n))})^{\log(n)}=(e^{\log(n)})^{\log(\log(n)}=n^{\log(\log(n)}