For which values of n does equation 2 become smaller than equation 1?

Oct 2012
3
0
Atlanta
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.
 

earboth

MHF Hall of Honor
Jan 2006
5,854
2,553
Germany
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.
 
Oct 2012
3
0
Atlanta
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
 
Last edited:

earboth

MHF Hall of Honor
Jan 2006
5,854
2,553
Germany
Thank you for this answer. Would this wolfram alpha plot accurately represent this solution?

...
1. Yes

2. I would use a slightly different scaling of the axes such that you can read the coordinates of the point of interception. (see attachment)
 

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Oct 2012
3
0
Atlanta
Thank you! Using a different scale definitely makes it more