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OpenStudy (kropot72):
\[\frac{1}{2^{n}}<0.01\]
Is this it?
OpenStudy (anonymous):
yes sir
OpenStudy (kropot72):
Do you want to take it step by step? The first step being to multiply both sides by 2^n.
OpenStudy (anonymous):
step by step will be more helpful
OpenStudy (kropot72):
Lets do it step by step. Can you multiply both sides by 2^n and post your result?
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OpenStudy (anonymous):
2^n < 0.01(2^n)
right?
OpenStudy (kropot72):
Not quite.
\[\frac{1}{2^{n}}\times\frac{2^{n}}{1}=?\]
OpenStudy (anonymous):
ok...
OpenStudy (anonymous):
then?
OpenStudy (anonymous):
1 < (0.01)(2^n)
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OpenStudy (kropot72):
Good work! The next step is to take logs of both sides of equation (1)
\[1<0.01\times2^{n}\ .........(1)\]
OpenStudy (anonymous):
log 1 < log (0.01)(2^n)
OpenStudy (anonymous):
0 < Log (0.01) + log 2^n
OpenStudy (kropot72):
Yes, but we can take the right hand side a step further:
\[\log\ 1<\log\ 0.01+n\ \log\ 2\ .......(2)\]
OpenStudy (anonymous):
got it... what will be the next step??
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OpenStudy (kropot72):
Subtract log 0.01 from both sides.
OpenStudy (anonymous):
2 < n log 2
OpenStudy (anonymous):
???
OpenStudy (kropot72):
You are correct. Now divide both sides by log 2.
OpenStudy (anonymous):
ok sir thanks a lot
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OpenStudy (kropot72):
You're welcome :)
OpenStudy (larseighner):
Of course if you take n to be an integer, as a child of the digital age you know the powers of 2 by heart. 128 (the 7th power) fits the bill so the 1/128 < .01. The sixth power doesn't.