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So I don't quite understand Shannons Law. Could somebody maybe give me an example and work it out? I have to do some equations using Shannons Law on a test soon without a calculator. Maybe I don't understand what log(sub2) and such is. Anyway the equation of shannons law is: $ C = B \cdot \log_{2}(1 + s/n) $ from my understanding $C =$ amount of data, $B =$ Bandwidth and $s/n =$ Signal to Noise ratio. I appreciate it guys.

NOTE I didn't know what tags to put on this, any help suggestions would be great.

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    @Howdy You still don't get it. $\text{Log}_{\text{sub}}$ is _not_ a way of writing $\text{sub}^y$, and $\log_2(1000)$ does not equal $10$ (though it is close: $\log_2(1024) = 10$ exactly).2011-11-02

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As J. M. says, $y=\log_2 x$ is the same as $2^y=x$. For a test it will probably help to know the powers of $2$, at least up to $2^{10}=1024$. This tells you that $\log_2 1024=10$

A discussion of the law is at Shannon-Hartley theorem. Intuitively, the bandwidth $B$ says we can transmit $B$ numbers per second. With very high signal to noise ratio we can resolve many levels in each number, so each number can carry many bits. If we can resolve $1024$ levels, we can get $10$ bits per number and carry $10B$ bits per second. If the channel is very noisy, we might need to send bits many times to get them through and the capacity goes down. If the signal to noise goes to zero, nothing should get through, and $\log_2 (1+0)=0$

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    You are right. But it would help to get a better feel for logs. $\log_2 512=9, \log_2 1000$ is very close to $10$2011-11-02