+0  
 
0
49
4
avatar+72 

If f(x) = (x+1)(x2+1)(x4+1)(x8+1)(x16+1)(x32​+1), then in simplified form f(2) = 2n-1. Determine the value of n.

MapleTheory  Dec 29, 2017
Sort: 

4+0 Answers

 #1
avatar+80973 
+2

If f(x) = (x+1)(x^2+1)(x^4+1)(x^8+1)(x^16+1)(x^32​+1), then in simplified form f(2) = 2n-1. Determine the value of n.

 

f(2)  = 

 

(2 + 1) (2^2 + 1) (2^4 + 1)(2^8 + 1) (2^16 + 1)(2^32 + 1)  =

 

(3)(5)(17)(257)(65537)(4294967297)  =

 

18446744073709551615  = 2^n  - 1       add 1 to both sides 

 

18446744073709551616  =  2^n        take the log of both sides        

 

log (18446744073709551615)  =  log (2)^n     and we can write

 

log (18446744073709551615)  = n* log (2)    divide both sides by log 2

 

log (18446744073709551615) / log (2)  =  n  =  64 

 

 

cool cool cool

CPhill  Dec 29, 2017
 #2
avatar+502 
+2

I got the same answer on a rough page but looking at the answer I was like 'w*f is this'

Rauhan  Dec 29, 2017
 #3
avatar+72 
+1

Wow thanks!!

MapleTheory  Dec 29, 2017
 #4
avatar+80973 
0

I actually think there is a more efficient way to do this........maybe someone else knows how

 

 

cool cool angel

CPhill  Dec 29, 2017

19 Online Users

avatar
We use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners.  See details