Given that log base2 5 = 2.3219, find an approximation for log base2 2.5?

Question

Given that log base2 5 = 2.3219, find an approximation for log base2 2.5
Can somebody tell me how to calculate this.

I am also having trouble with a word logarithm problem:
If a telephone network is designed to carry C telephone calls simultaneously, then the number of switches needed per call must be at least logbase2 C. If the network can carry 10 000 calls simultaneously, how many switches would be needed for one call and for 10 000 simultaneous calls?

Answer 1

log(base2)2.5 = log(base2)(1/2)(5).........use properties of logs to expand right side



= log(base2)(1/2) + log(base2) 5..........but log(base2)(1/2) = log(base2)(2^-1) = -1



= -1 + 2.3219 = 1.3219................and you're done.





For the second: if C = 1, log(base2) 1 = 0; so no switches are required for one call.



log(base2)10,000 = log(10,000)/log(2) (change of base) = 4/log(2) = 13.3........so 14 switches.

Answer 2

log (base 2) 2.5

= log (base 2) (5/2)

= log (base 2) 5 - log (base2) 2 { log (base n) n=1 }----- formula

=2.3219 -1= 1.3219





answer of the first question.



part b) log (base 2) 10,000 =

log (base 2) 2*5*2*5*2*5*2*5

= log (base 2) (2^4)+ log (base 2) (5^4)

=4 log (base 2) 2 + 4log(base 2) 5

= 4*1 +4*(2.3219)

=4+9.2876

= 13.2876



so, at least 14 switches needed to carry 10,000 simultaneous calls.

Answer 3

Log_2 2.5 = Log_2 5/2 = Log_2 5- Log_2 2 = 2.3219 -1 = 1.3219

Answer 4

the respond is C. I used my epic ti-89 titanium graphing calculator that i take advantage of for my very own pre-calculus class, and it solved the equation. in order that which you do no longer finally end up counting on your calculator each and all the time for each little thing, this is what's actual being carried out: a*log[ok](b) the place [ok] is the backside in the logarithm of (b). So, you will desire to get it so each and all the areas of your equation are in a sort basically like the single above different than without the "a" in front of them: a*log[ok](b) = log[ok](b^a) = log[ok](x) then you definitely basically plug in the numbers: 2*log[2](25) = log[2](25^2) = log[2](625) and 2*log[2](5) = log[2](5^2) = log[2](25) then you definitely basically would desire to do the maths: log[2](625) - log[2](25) + log[2](3) = (625/25) * 3 = seventy 5 = log[2](seventy 5) i wish this facilitates.

Answer 5

log_2 5 = 2.3219



log_2 2.5 = log_2 5 - log_2 2 = 2.3219 - 1 = 1.3219

Answer 6

ld2.5 = ld(5/2) = ld5-ld2 = 2.3219-1 = 1.3219

Answer 7

http://www.flickr.com/photos/dwread/5592477556/in/stream

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No switches needed for 1 call.

14 switches needed for 10000 calls.

http://www.flickr.com/photos/dwread/5592492600/in/stream