No, log3 to the base 2 is irrational.
b = 6
by transitive property
a=b=3.60555
12
No, log3 to the base 2 is irrational.
log AB^2 log A+log B+log2
C = 2*B*log2(M) where C --> capacity B --> bandwidth M --> # of discrete signals
Use the LOG function. =LOG(n,b) n = Number b = Base =LOG(2,10) = 0.30103
The expression x = log3(100) is equivalent to 3x = 100. One way to calculate logs with a base of (b) is: logb(y) = log(y) / log(b). So in this case, you would have log(100) / log(3) = 4.192 [rounded to 3 decimal places]
if A & B matrices then A into B not equals to B into A
Nyquist has shown that C=2 B log2 (M) [bps] C: Capacity B: Bandwidth M: Signaling Levels log2 (x)= [ log10 (x) / log10 (2) ] So We assume F1=0Hz F2=20kHz so B= F2-F1=20000 Hz C = 2*20000* log2 (16) = 2*20000* log10 (16) / log10 (2) = 160 000 bps
Using the Shanon Capacity formula,C = B log2 (1+ SNR)Where B = 20 X 106 HzIn order to calculate C (channel capacity) one must be mindful that 1Kbps = 1024 bps, NOT 1000 bps. Having this in mind,C = 100 X 1.024 * 106 bps = 1.024 * 108Substituting C, B, and solving for SNR:1.024 * 108 = 20* 106 * log2 (1+SNR)1.024 * 102 = 20 * log2 (1+SNR)5.12 = log2 (1+SNR)25.12 = 1 + SNR2 5.12 - 1= SNRSNR = 33.8if in decibels = 10 log (33.8) = 15.29 dB
2a. (a, b and c are all equal.)
If a equals 3 and b equals minus 5 then a minus b equals what
6 === the what does B =
C = B * log2(1 + SNR) C= channel capacity B= Bandwidth , telephone lines have a usable range of around 3400Hz = =