Determining the value of n in inequality with logs I have two functions $f(n) = n^{1.001}$ and $g(n) = nlog_2n$ and I want to determine if the inequality $f(n) < c*g(n)$ is true, where $f(n), g(n), c > 0$.
To do this, I am using the following property:
$f(n) < c*g(n)$ if and only if $log(f(n)) < logc + log(g(n))$
So:
$1.001*log_2n < log_2c + log_2{(nlog_2n)}$ 
By setting c to 1:
$1.001*log_2n < log_2n(nlog_2n)$
I am not sure how can I solve this inequality so that I can find what the value of 'n' is.

\begin{align}
\lim_{n \to \infty} \frac{f(n)}{g(n)} &=\lim_{n \to \infty} \frac{n^{1.001}}{n \log_2 n} \\
&= \lim_{n \to \infty} \frac{n^{0.001}}{\log_2 n} \\
&= \ln 2 \lim_{n \to \infty} \frac{0.001 n^{-0.999}}{\frac1n} \\
&= 0.001 \ln 2 \lim_{n \to \infty } n^{0.001} \\
&= \infty
\end{align}
Hence $\frac{f(n)}{g(n)}$ can't be bounded by a constant.
Also for your question in the comment, yes, $\lim_{n \to \infty}\frac{g(n)}{f(n)}=0$
