Question: Here is the question repeated: Arrange the following in order into increasing order of growth rates. $$(1.5)^n, n^{100}, (\log n)^3, \sqrt n\log n, 10^n, (n!)^2, n^{99}+n^{98}, 101^{16}$$

I graphed these functions in my calculator and think that this is the correct ordering:$$101^{16},(\log n)^3\sqrt n\log n, n^{99}+n^{98}, n^{100}, (1.5)^n, 10^n, (n!)^2$$ 

(I only used my calculator for the log graphs.)

Answer: 
Your answer is correct because
$101^{16}$ is fixed in terms of growing.
$n^{100}$ is slower than $1.5^{n}$.
And $(n!)^{2}$ is significantly large.
Your answer is correct. To compare the logarithm you should notice that 
$$\lim\_{x->\infty}\frac{(\log(x))^p}{x}=0$$ 
for any positive constant $p$ and from this you immediately see that $(\log n)^3$ has smaller rate that $\sqrt n\log n$ and both have smaller order than all polynomials.