# How can I quantify the amount of space required to store all possible 128kilobit mp3s?

Within, say, a collection of every possible 30 second long MP3 file encoded at 128kbps, I'd probably be infringing on a few thousand copyrighted works.

128kilobits per second = 128,000 bits per second * 30 seconds = 3,840,000 bits.

There are 2 to the 2,840,000 possible files of that length.

Ignoring the fact that most of those won't be valid mp3s, how can I quantify the amount of space needed to store all those files? For instance, is that more bits than there are atoms in the universe?

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According to Wikipedia (en.wikipedia.org/wiki/Observable_universe#Matter_content) there are about $10^{80}$ atoms in the universe. So yes, that is far, far more bits than there are atoms! –  Brad Oct 30 '12 at 12:28
@Brad - I'll admit it: my math skills are terrible. How do I put 2 to the X in terms of 10 to the Y so I can compare? –  Nathan Long Oct 30 '12 at 12:34
Very crudely: 2^2,840,000 = (2^4)^(2,840,000/4) = 16^710,000 > 10^710,000 > 10^80. –  Marcks Thomas Oct 30 '12 at 12:47
For comparison, we generate about 3 quintillion bytes (18 zeros) of data per day on the Earth compared to ~10^80 atoms in the universe. –  Amzoti Oct 30 '12 at 12:55
The first equation has the units incorrect. It is not 128,000 bits, it is 128,000 bits/second. The error is corrected across the second equal sign. Then a typo in the next line. –  Ross Millikan Oct 30 '12 at 13:06

The total space is $2^{3,840,000}\cdot 3,840,000$ bits. This is a fine quantification, but if you want the number of zeros in base $10$, you can take the log of it. $$\log_{10}(2^{3,840,000}\cdot 3,840,000)=3,840,000 \log_{10} 2 + \log_{10} 3,840,000 \\ \approx 3,840,000\cdot 0.30103 + 6.58433 \approx 1,155,961$$