+0

0
93
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+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
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#1
+82886
+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

CPhill  Dec 29, 2017
#2
+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
+72
+1

Wow thanks!!

MapleTheory  Dec 29, 2017
#4
+82886
0

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

CPhill  Dec 29, 2017

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