yes, even if we take $n=2^{x}$ , we get same result.
but in case of log, we need to more derive it.
It means $n=2^{x}$, i.e. exponential growth rate function shows growth rate more accurately
There $\frac{n}{\log n}=\frac{2^{x}}{x}$ , which is higher growth rate than $n^{\frac{1}{3}}=2^{\frac{x}{3}}$
right??
But in case of taking log , it's rate is less.
So, this function works fine in higher range.
Log function works well, if two points of a function in less range works suppse between $\left ( 5,6 \right )$, they have massive growth.
right??