robinfr93
if 2^n1 > 1000
Find the least value of n.



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sauravshakya
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2^n>1001
log2^n>1001
nlog2>1001
n>1001/log2

sauravshakya
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Does it help?

lgbasallote
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i think that's 2^(n1)....

robinfr93
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\[2^{n1} > 1000\]
Find the least value of n

Yahoo!
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yes...i think lg is correct

sauravshakya
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Oh..

robinfr93
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answer according to the text is 11.

robinfr93
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i.e, n = 11

sauravshakya
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2^(n1)>1000
log2^(n1)>log1000
(n1)log2>log1000
n1>log1000/log2
n1>9.965
n>10.965

robinfr93
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thanx :D

sauravshakya
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Welcomx

UnkleRhaukus
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\[2^{10}=1024\]
\[2^{111}=1024\]