I have to prove or disprove the following: $$ 2xlog_{10}((x+2)^2) + (x+2)^2log_{10}(\frac x2) \in O(x^2log_{10}(x))$$
My approach (with $log$ is meant $log_{10}$):
$4x log(x+2) + (x+2)^2log(x) - (x+2)^2log(2) \le c x^2 log(x)$
$4xlog(x) + (x+2)^2 log(x) - (x+2)^2 log(x) \le c x^2log(x)$
$4 \le cx$
For example for $c=1$ and $x \ge 4$ the inequality is correct. But if I plot it in WolframAlpha, I can see it is obviously not correct. Where's the mistake? WolframAlpha Plot
On
Ok, I fear I have now to give it a try. The problem is that solving this requires some understanding of the meaning of the Landau symbols, but I try to make it as elementary as possible avoiding all order manipulation of the terms. First, it does not matter if one uses the $\ln$ or the $\log_{10}$ for asymptotic estimates here, since changing from one to the other means only multiplying by a constant, which does not change the asymptotic order.
One has: $$ 4x\ln(x+2) = o(x^2\ln(x))$$ $$ (x+2)^2\ln(x/2)=O(x^2\ln(x))$$ This can be shown using the rule of l'Hospital.
So the first term of the lhs is negligible compared with the rhs and the second term is of the same order. Therefore: $$4x\ln(x+2)+(x+2)^2\ln(x/2)=O(x^2\ln(x)) $$
In order to prove the asymptotic relationship $f \in O(g)$ you have to choose both constant $c$ and $x_0$, such that the inequality $f(x) \le c \cdot g(x)$ becomes true for all $x > x_0$.
In your case you can choose $c = 8$ and $x_0 = 4$. I'll leave all the rest to you. Just a hint - you can replace $x + 2$ by $2 \cdot x$ in this case (for simplicity, because $x + 2 \le 2 \cdot x$) and do all such minor simplifications to reduce your $f(x)$ to a most simple form.
(By the way, it's a usual practice to use $n$ in such expressions)