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Mathematics 12 Online
OpenStudy (anonymous):

1/2^n < 0.01 what is the value of n

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?

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)

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??

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

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.

OpenStudy (kropot72):

\[n >\frac{2}{\log\ 2}=6.644\]

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