anonymous
  • anonymous
if 2^n-1 > 1000 Find the least value of n.
Mathematics
chestercat
  • chestercat
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anonymous
  • anonymous
2^n>1001 log2^n>1001 nlog2>1001 n>1001/log2
anonymous
  • anonymous
Does it help?
lgbasallote
  • lgbasallote
i think that's 2^(n-1)....

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anonymous
  • anonymous
\[2^{n-1} > 1000\] Find the least value of n
anonymous
  • anonymous
yes...i think lg is correct
anonymous
  • anonymous
Oh..
anonymous
  • anonymous
answer according to the text is 11.
anonymous
  • anonymous
i.e, n = 11
anonymous
  • anonymous
2^(n-1)>1000 log2^(n-1)>log1000 (n-1)log2>log1000 n-1>log1000/log2 n-1>9.965 n>10.965
anonymous
  • anonymous
thanx :D
anonymous
  • anonymous
Welcomx
UnkleRhaukus
  • UnkleRhaukus
\[2^{10}=1024\] \[2^{11-1}=1024\]

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