Please allow me to introduce myself; Name is Morris, I am a computer programmer.
For which values of n does equation 2 become smaller than equation 1?
1. 8(n^2)
2. 64 n (log n to base 2)
Show clear working please.
1. I assume that you want to solve for n the inequality
\(\displaystyle 8n^2 < 64 n \log_2(n)\)
2. What kind of number is n? Real or natural? I'm going to use real numbers!
3. It is a little bit easier to solve the equation
\(\displaystyle 8n^2 = 64 n \log_2(n)\)
for n, but I doubt that you can solve this equation algebraically. So use a numerical iterative method.
With Newton's method you'll get \(\displaystyle n \approx 1.099997\)
4. To answer your question: For \(\displaystyle n \in (0, 1.099997)\) the 2nd term is smaller than the 1st one.