For each of series find the smallest $k$, that $a_n = O(n^k)$ Hey I need you to check my solutions:
a) $a_n = (2n^{81.2}+3n^{45.1})/(4n^{23.3}+5n^{11.3})$
This one is done from $\sum_{i=1}^{k} O(a_i(n)) = O(max\lbrace a_i,..,a_k \rbrace )$
So it's $n^{81.2}/n^{23.3} = n ^ {57.9}$, thus $k = 57.9$
b) $a_n =5^{\log_2(n)} $ so I have $5^{\log_2(n)} \le Cn^k \iff 5^{\frac{1}{log_n(2)}} \le Cn^k \iff 5 \le C(n^{log_n(2)})^k \iff 5 \le C \cdot 2^k$ and now im not really shure which $k$ is the smallest. $ k = 0 $ ?
c) $a_n = (1.001)^n$,  so equation is $ (1.001)^n \le C \cdot n^k \iff log_n(1.001)^n \le C \cdot log_n(n^k) \iff nlog_n(1.001) \le C \cdot k \iff 0 \le -log_n(1.001)n + C \cdot k$  And now I've not really shure. Wolfram alpha shows $k \ge \frac {0.009995n}{ln(n)}$ and $ \lim_{n \to \infty} \frac {0.009995n}{ln(n)} = 0$, so maybe $k = 0$?
d) $a_n = nlog^3(n) $ And about this one I've no idea.
So if somebody can please check my $a,b,c$ solutions and give some hints on $d)$. Thanks in advance.

b) $5^{\log_2n}=2^{\log_25\log_2n}=(2^{\log_2n})^{\log_25}=n^{\log_25}$
c) Try to show that no matter what $k$ is, $a_n$ is not $O(n^k)$. 
d) There is no smallest $k$, but try to show that $k=1$ doesn't work, but that for every positive $\epsilon$, $k=1+\epsilon$ works. 
