How do you realize that $$(3log^2(n) + 55log(n^{10})+8log(n))*log(n) \neq \Omega(log^{10}(n))$$ ,where $log^x(n)$ means $(log(n))^x$
I know that by definition, if $f(n) = \Omega(g(n))$ then there exists a constant $c_1$ such that $$0 \leq c_1 *g(n) \leq f(n)$$ for all $ n > n_0$. In our case it yields $$c_1*log^{10}(n) \leq (3log^2(n) + 55log(n^{10})+8log(n))*log(n) $$
$\Rightarrow$
$$c_1 \leq \frac{3log^3(n)}{log^{10}(n)} + \frac{55log(n^{10})*log(n)}{log^{10}(n)} + \frac{8log^{2}(n)}{log^{10}(n)}$$
RHS: It's fairly easy to see that the denominator is larger than the nominator of both the first and third term, which means that $55log(n^{10})*log(n) > log^{10}(n) + c_1$ for some $n_0$ if the first equation holds.
I am just not able to see how to manipulate the equation to realize it...
I'll take it from where you paused. Assume your claim is true, that is the following is correct $$(3\log^2(n) + 55\log(n^{10})+8\log(n))\log(n) \neq \Omega(\log^{10}(n))$$ Then there exists $c_1 > 0 $ and $n_0 > 1$ such that for all $n > n_0$,we have $$c_1\log^{10}(n) \leq (3\log^2(n) + 55\log(n^{10})+8\log(n))\log(n) $$ Divide both sides by $\log(n)$ $$c_1\log^{9}(n) \leq (3\log^2(n) + 55\log(n^{10})+8\log(n)) $$ Use $\log a^b = b \log a$ $$c_1\log^{9}(n) \leq (3\log^2(n) + 550\log(n)+8\log(n)) $$ Divide again by $\log (n)$ $$c_1\log^{8}(n) \leq (3\log(n) + 558) \tag{1}$$ Your task now is to find me a $c_1 > 0$ and $n_0 > 1$, such that for all $n > n_0$ such that $(1)$ is true. Written differently, $$\exists n_0 > 1 \quad \mid \quad 0<c_1\leq (\frac{3}{\log^{7}(n) } + \frac{558}{\log^8(n)}), \ \forall n \geq n_0$$ The above is clearly not true. The larger $n$ is, the upper bound goes to zero. By the sandwich theorem, you get that $c_1$ has to be zero. By contradiction, the initial hypothesis is false.