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.
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.
Thank you for this answer. Would this wolfram alpha plot accurately represent this solution?
http://www.wolframalpha.com/input/?i=plot+8+x+^+2+%2C+64+x+log+[2%2Cx]%2C+x%3D0+to+2
In Wolfram alpha : plot 8 x ^ 2 , 64 x log [2,x], x=0 to 2