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Title: Can someone explain how to solve this Logarithm word problem? Post by: lhiggin on Aug 24, 2012 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? Title: Can someone explain how to solve this Logarithm word problem? Post by: c.c.morgan on Aug 24, 2012 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.
_ Title: Can someone explain how to solve this Logarithm word problem? Post by: iliya1 on Aug 24, 2012 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. Title: Can someone explain how to solve this Logarithm word problem? Post by: onestep on Aug 24, 2012 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 รท In 2 = 13.29, or 14 Title: Can someone explain how to solve this Logarithm word problem? Post by: ilindsey05 on Aug 24, 2012 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 |