I need to solve some logarithm equations and unfortunately I've been trying for a while to solve it but without success.
The problems are:
1) For what values of N is $10NlgN > 2N^2$
My Answer: $10lgN > 2N$ ==> $lgN^{10}$ > $2N$ ==> $N^{10} > 2^{2N}$
And I do not know hot to continue
2) For what values of $N$ is $N^{3/2}$ between $N(lgN)^2/2$ and $2N(lgN)^2$
lgN = binary logarithm (base 2)
Many thanks :)
On
For the first part, consider the function $$f(x)=\frac{10 }{\log (2)}\log (x)-2 x$$ $$f'(x)=\frac{10 }{\log (2)}\frac 1x-2$$ $$f''(x)=-\frac{10 }{\log (2)}\frac 1{x^2}$$ The first derivative cancels for $x_*=\frac{5}{\log (2)}$ and $$f(x_*)=-\frac{10 \left(1+\log \left(\frac{\log (2)}{5}\right)\right)}{\log (2)}\approx 14.08$$ and the second derivative test shows that this is a maximum.
So, there are two solutions $x_{1,2}$ for which $f(x)=0$ and $f(x)$ is positive betwenn these two solutions.
Sooner or later, you will learn that any equation which can write $$A+B x+C\log(D+Ex)$$ has solution(s) which can express in terms of Lambert function. Applied to this case, you would get $$x_1=-\frac{5}{\log (2)} W\left(-\frac{\log (2)}{5}\right)$$ $$x_2=-\frac{5 }{\log (2)}W_{-1}\left(-\frac{\log (2)}{5}\right)$$ The wikipedia page provides series expansions which allow the computation of Lambert function for small arguments. Using them, you should get $x_1\approx 1.18$ and $x_2=22.44$.
So, back to $N$, $10N\log_2(N)>2N^2$ when $2 \leq N \leq 22$.
Look along powers of 2 to find 2 powers where the inequality switches, since it always will fail for large $N,$ the log function is dominated by all polynomials. Then numerical binary search will do it. This will work for both, I demonstrate one.
1) $\lg N>N/5$: $N=2^4$ gives $4>16/5$, which is fine but $N=2^5$ gives $5>32/5$ which is false. So it is some $N \in (16,32).$