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if 2^n-1 > 1000 Find the least value of n.

Mathematics
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2^n>1001 log2^n>1001 nlog2>1001 n>1001/log2
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i think that's 2^(n-1)....

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

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