I am doing an order of growth question that involves logarithms. I have an expression and I need to determine which function controls the growth. I have come across a question that I cannot see how they got the order of growth to be that answer. The question is below.
$$\text{Find the order of growth of }2^{\log(n^2)} + 2^{\log(n)^2}.$$
They are saying that the order of growth is $n^{log(n)}$. I don't know how they got this answer. Any help would be appreciated.
$$y = 2^{\log(n^2)} + 2^{\log(n)^2}$$ $$y = (2^{\log(n)})^2 + (2^{\log(n)})^{log(n)}$$
Substitute $log(n) = \frac{log_2(n)}{log_2{10}}$ and for less cluttering I'm replacing $\frac{1}{log_2{10}}$ by $k$.
Hence, $log(n) = k.log_2{n}$
$$y = (2^{\log_2(n)})^{2k} + (2^{\log_2(n)})^{k.{log(n)}}$$ $$y = n^{2k} + n^{k.{log(n)}}$$
Substitute $k$ where $k = \frac{1}{log_2{10}} = 0.30111$
$$y = n^{0.60222} + n^{0.30111.{log(n)}}$$
Now I think you can calculate the growth from reduced expression.