How to solve the following recurrence relation? I get confused when decimals are used in the expression.

Question

$T\left ( n \right )= 8T\left ( \frac{n}{2} \right )+\left ( n\cdot logn \right )^{2.99}$

Also can $\mathcal{O}(n^{3})$ be an upper bound to above recurrence relation?

Comments

you can use master theorem here.   a=8 b=2   k =2.99 and p = 2.99

a>b$^{k}$    (8>2$^{2.99}$)   and if p>=0  then n$^{k}$ log$^{p}$n.

so n$^{2.99}$log$^{2.99}$n

let  n$^{2.99}$log$^{2.99}$n   = X

we can easily see  X<n$^{3}$.  so yes it is upper bond .