We can assume that $m$, and $n$ are reals greater than $1$. I'm hoping that we can find a (good) bounding function for $m$, meaning a function like:
$$m > f(n)$$
from the inequality
$$2n> (\log{(n^{10} m)})^{m}$$
In other words, I'm trying to find how large $m$ can be for a given $n$. Please note that I'm looking for someone to derive the solution. In other words, derive the solution that you get.
I've tried using Mathematica to solve this, but it doesn't work. I've been trying to find a way to collect the $m$ terms together, but so far I've had no luck. Can someone please help me to solve this problem?
Assuming that you want to solve for $m$ the equation $$2n=\log{\big((n^{10} m)}^{m}\big)=m\left(\log(m)+\log(n^{10})\right)$$ the solution is given in terms of Lambert function $$m=\frac{2 n}{W\left(2 n^{11}\right)}$$ and, for large values of $x$, you can use the approximation $$W(x)\approx L_1-L_2+\frac{L_2}{L_1}+\frac{L_2(L_2-2)}{2L_1^2}+\frac{L_2(2L_2^2-9L_2+6)}{6L_1^3}+\cdots$$ where $L_1=\log(x)$ and $L_2=\log(L_1)$.
So, a first approximation could be $$m\sim \frac{2 n}{\log \left(n^{11}\right)-\log \left(\log \left(n^{11}\right)\right)}$$
Trying for $n=10^6$, the approximation gives $m=13610$ while the exact solution would be $m=13544$.
Edit
Obviously, I misread the equation which is now $$2n=\left(\log(n^{10} m)\right)^m$$ for which, I am afraid, only numerical methods could work. Since $m$ vary very slowly with $n$, let $\color{red}{n=10^k}$ and use Newton method to get the following table $$\left( \begin{array}{cc} k & m \\ 1 & 0.95568 \\ 2 & 1.38094 \\ 3 & 1.79113 \\ 4 & 2.18554 \\ 5 & 2.56742 \\ 6 & 2.93927 \\ 7 & 3.30285 \\ 8 & 3.65950 \\ 9 & 4.01017 \\ 10 & 4.35565 \\ 11 & 4.69653 \\ 12 & 5.03330 \\ 13 & 5.36637 \\ 14 & 5.69608 \\ 15 & 6.02270 \\ 16 & 6.34650 \\ 17 & 6.66767 \\ 18 & 6.98640 \\ 19 & 7.30287 \\ 20 & 7.61720 \\ 21 & 7.92953 \\ 22 & 8.23998 \\ 23 & 8.54864 \\ 24 & 8.85562 \\ 25 & 9.16099 \\ 26 & 9.46483 \\ 27 & 9.76722 \\ 28 & 10.0682 \\ 29 & 10.3679 \\ 30 & 10.6663 \end{array} \right)$$
A quick and dirty regression of the totally empirical model $$m=a+b k^c$$ seems to be very good $(R^2 \sim 1)$ $$\begin{array}{clclclclc} \text{} & \text{Estimate} & \text{Standard Error} & \text{Confidence Interval} \\ a & 0.415226 & 0.002492 & \{0.410114,0.420338\} \\ b & 0.530714 & 0.000956 & \{0.528752,0.532675\} \\ c & 0.870543 & 0.000484 & \{0.869551,0.871536\} \\ \end{array}$$