Q:

Finding asymptotic bounds for $T(n)=16n^4T(n^{\frac{1}{2}})+2n^8\log^4n$

I managed to get something after some work but it doesn't feel right. 
$\log$ means logarithm with base $2$.
Here is what I did:
$T(n)=16n^4T(n^{\frac{1}{2}})+2n^8\log^4n$
Substitute $m=n^4$:
$T(m^{\frac{1}{4}})=16mT(m^{\frac{1}{8}})+\frac{1}{128}m^2\log^4m$
Defining $S(m)=T(m^{\frac{1}{4}})$:
$S(m)=16mS(m^{\frac{1}{2}})+\frac{1}{128}m^2\log^4m$
Dividing by $m^2$ and defining $U(m)=\frac{S(m)}{m^2}$:
$U(m)=16U(m^{\frac{1}{2}})+\frac{1}{128}\log^4m$
It is rather easy to show that (hopefully this is correct): $U(m)=\theta(\log\log m*log^4m)$
That means that $S(m)=\theta(m^2*\log\log m*\log^4m)$
From here $T(m)=\theta(m^8*\log\log m*\log^4m)$
And finally $T(n)=\theta(n^{32}*\log\log n*\log^4n)$
I am sure there are simpler ways, but my main question is if what I did is correct, and if I made a mistake, where is it?

A:

I agree with you up to the formula for $T(m)$.  But I don't think you get an extra fourth power by changing letters.  You already did the fourth power when you changed from $S$ to $T$.

