# 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.

• It's not true (no matter the $c$), and the reason is simple. $f(n)/g(n)\to \infty$ when $n\to \infty$ Mar 11 '19 at 22:45
• Would it be correct to say that g(n)/f(n) → 0 when 𝑛→∞? Since f(n) grows faster than g(n)? Mar 12 '19 at 0:43

\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, $$\lim_{n \to \infty}\frac{g(n)}{f(n)}=0$$