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Home Q & A Board Science-Related Homework Help Mathematics
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lhiggin lhiggin
wrote...
12 years ago
I am taking a Math class and I haven't been doing this type of math for a few years, so a well explained answer would be great.

If a telephone network is designed to carry C telephone calls simultaneously, then the number of switches needed per call must be at least log2 C. If the network can carry 10,000 calls simultaneously, how many switches would be needed for one call and for 10,000 calls?
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wrote...
#1
12 years ago
2^14 is a little more than 16,000, so log base 2 of 10,000 is around 13.3 or so.  They'd need 14 switches for 10,000, 0 for 1.
_
C.C.Morgan
iliya1 Author
wrote...
#2
12 years ago
C = 1, Sw = log(base 2) 1 = 0. (But that may be wrong in the real world -- a degenerate case -- since there may be a minimum of 1 switch to activate the call and hang it up. )

C=10,000, Sw = log (2) 10000 = ln 10000/ln 2 = 13.2877 __> 14 rounded up.
wrote...
#3
12 years ago
log(2) (10,000) = number of switches

then, by the definition of a logarithm, we can write:

2^(number of switches) = 10,000

(number of switches)*In 2 = In10,000 . . . taking natural log of both sides

number of switches = In10,000 ÷ In 2

= 13.29, or 14
wrote...
#4
12 years ago
log (xy) =log (x) + log (y)
log (2*C) = log 2 + log C.
if u wanna mention base as 2, log2 C
 
then logx (y)= logm (y)/logm (x)    [m, is any number 10,e etc]
      log2 (C)= log10 (C)/log10 (2)
log10 is log in calculators.
loge is "ln" in calculators

for 2nd procedure ans is 13.2877 ~14 thanks
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