GT=t*log2/logb-logB\n\nif i have the data of OD for logb and OD for logB, and i know the duration between b and B..so i shall be able to get the generation time, right...my teacher says it has to be the case that B is twice of b...does it have to be? .....(i am confused..am i making any sense..)
216289
Actually, your teacher is wrong.  It all depends on the growth phase of the bacterial culture.   There are 3 phases in the growth of a bacterial culture; the lag phase, the log phase, and the stationary phase.  The lag phase is exactly what it describes...the bacteria divide veeeeerrrryyy slowly.  In simple terms, this is mainly due to low amounts of certain "growth signals".  It usually takes bacteria several hours to get out of lag phase and into log phase.  \n\nLog phase is the stage where the bacteria are dividing at a constant rate.  This is when you're going to take measurements for the growth rate.  Say, for example, that your cells have a doubling time of 1hr.  So if your beginning OD is 0.8 your end OD after an hour would be 1.6.  In this case B is twice b.  However using the same example, if we grow for 2 hours instead of 1, you've gone thorugh 2 doubling times and the final OD would be 3.2 verses 0.8 (B is 4x b).  So the relation between b and B is strictly dependant on the time (which is why "t" is a component of the equation) and growth rate of the cells.\n\nThe final phase of bacterial growth is the stationary phase.  This is where the number of bacteria increases to such a high density that further division is inhibited (by physical limitations such as nutrients, pH, cell/cell inhibition, etc) and the growth substatially slows.\n\nI hope this clearly answers your question.
203732
Okay, this is how I read this problem:\n\nThe integral of 1/(t-1) dt on the interval from 0 to (1-3x+x^2).\n\nFirst look at what you're integrating 1/(t-1) dt.  This can be rewritten (t-1)^-1 dt.  If you substitute u for t-1 and du for dt (the derivative of t-1 = 1) then you have the integral of u^-1 du.  This is ln |u| (ln is the natural log).  Sub t-1 for u and you get ln |t-1|.\n\nNow you use your interval and you get:\n\nln |(1-3x+x^2) - 1| - ln |0|\n\nSimplify and you get:\n\nln |x^2 - 3x| - ln |0|
499233
T(n) - T(n-1) = log(n).\nT(n-1) - T(n-2) = log(n-1)\nT(n-2) - T(n-3) = log(n-2)\nT(n-3) - T(n-4) = log(n-3)\n...\nT(2) - T(1) = log(2)\n\nThe sum of all of these equations is\nT(n) - T(1) = log(n) + log(n-1) + ... + log(2) = log(n!)\nIsn't it great how nearly all of the terms on the left side cancel and all of the terms on the right collapse into one simple term?\nT(n) = log(n!) + T(1)\nT(n) = log(n!) + C
402804
OK, there's 2 possibilities:\nIf you meant:\nlog (16^1/4), then you need to know the property of logarithms where:\nlog a^b = b log a.\n\nSo:\nlog 16^(1/4) = 1/4 log 16\nlog 10^(.01) = .01 log 10\nlog 2^(1/8) = 1/8 log 2\n\nIf you meant (log 16)^1/4, then I have no idea how to do it by hand.\n\nEdit:\nFurther, you can also do things like:\nlog 16 = log 2^4 = 4 log 2\nso 1/4 log 16 = 1/4 * 4 log 2 = log 2.
541512
Type 1.16 log / 4 / 1.0125 log =\nThis is 3 approximately (2.9869...)\n\nTh.
373702
Do you know your exponent rules? The same apply for logs. Multiplication means addition (this is why x^5 times x^2 is x^7 and NOT x^10) and division means subtraction (this is why x^16 divided by x^4 is x^12 NOT x^4).\n\nKeeping these rules in mind, I'll show you some examples of how these rules can be applied.\nlog x + log y = logxy\nlog 2x + log 4x = log8x^2\nlog12 = log6 + log2\nlog 4x - log 9x = log4x/9x\nlog2 - log4 = log2/4\nlog10x/3x = log10x - log 3x\n\n\nYou can also pull out numbers to the front of the logs like so:\nlog10x = 2log5x\nRemember with this rule: you can only pull NUMBERs out, NOT variables!\nSome more examples of this:\n4log4x = log16x\nlog45x = 9log5x\n2log6x = log12x\n\n\nNow I'll give you an example of all of them together:\n2log5x + 2log6x\nlog10x + log12x\nlog120x^2\n\nAnother example:\n4log4x - 2log2x\nlog16x - log4x\nlog16x/4x
40325
In the army we always started our logs @ 0000 and closed them out at 2400. so i'm not sure where your getting your info. the time 24:00:00 does, after all, only exist for one second and then becomes 00:00:01
338546
Logs are just equivalent to power statements. \nlog_a (b) = c means a^c = b.\nSo that equation just means x^2 = 2x-3.\nNow thats just a quadratic;\nx^2 - 2x + 3 = 0\nHowever that has no real roots. So there are no such values of x. But hopefully you get the idea.
620020
No one on here is going to actually DO your homework for you.. but here is a place for you the get help -\n\nhttp://www.themathpage.com/aPreCalc/logarithms.htm\n\nhttp://en.wikipedia.org/wiki/Logarithm\n\nhttp://physics.mtsu.edu/~wmr/log_2.htm
19743
UGH!!!  Thx for the effort (I was the one to post that question, as a part of a solution to another problem I found here on Yahoo! Answers.)  Incidentally, I made a mistake when simplifying the other problem and it should be xln(2)-2ln(x)-ln(86400) = 0\n\nI don't know any way to get rid of the ln(x) unless maybe it's using calculus techniques?  I'm just glad this problem is making other ppl trouble about it as much as I did :-).
230456
well chinaren got it started but I believe his explianation is weird.\nLets assume x=3^t\nthen log3(x)+log2(x+1)=3\n=>log3(3^t)+log2(3^t+1)=3\n=>tlog3(3)+log2(3^t+1)=3\n=>log2(3^t+1)=3-t\n=>3^t+1=2^(3-t)\n=>3^t+1=2^3/2^t\n=>(3^t+1)*2^t=2^3\n=>3^t*2^t+2^t=8\n=>6^t+2^t=8\nfrom here there is one one solution possible as t=1\nso x=3^1=3
54147
For any AP,\n \nNth term = A + (n-1)d\nwhere A is first term, d is common difference.\n\n A = log (a)\n\n d = difference between any two consecutive terms\n   =  log(ab) - log(a)\n   =  log(ab/a)     ...... property of logarithm\n   =  log b\n\nHence, nth term is :\n\n T(n) = log(a) + (n-1)log(b)\n\nThat's your answer. I hope this solves your problem.
287902
(log3)25 = 2 * (log3)5\n\nSo, it seems that:\n\n2 * ((log3)5) / (log3)5 =\n2 * ((log3)5 / (log3)5) =\n2 * 1 =\n2.\n\nDon't quote me on that.
348499
1)\nlog(base6)(40)= log(base10)(40)/log(base10)(6)\n\n2)  Factor 42 to  7*3*2.\n\n  ln(42)=ln(7*3*2)=ln7+ln3 + ln2\n\nFactor 45 to 3*3*5 .\n\n  log45=log(3*3*5)=2log3+log5
643245
the change of base rule says that\n\nlog(a)x = log(b)x / log(b)a\n\n(the (a) and (b) represent the base of the log).  Since the Common log is base 10, what then (b) in the formula is 10.\n\nlog(a)x = log x / log a\n\nFor example, if you want to evaluate log(2)5, you would do log 5/log 2.
85019
Since the u235 decays 6000 times as fast as the u238 (4470/.7), if we tried graphing the decay equations on a graphing calculator, the u238 "curve" would look like a horizontal line.  So we just focus on the decay of the u235.  The basic equation would be 0.0072 = (0.5)^(t/0.7), with t measured in billions of years.  When t = 0.7 billion = 700 million, (0.5)^(0.7/0.7)=(0.5)^1 = 0.5, that is half of what you started with.\n\nTo solve the equation, take the log of both sides.\n\nlog(0.0072)=log[(0.5)^(t/0.7)]\nlog(0.0072)=(t/0.7)*log(0.5)\nlog(0.0072)/log(0.5)=t/(0.7)\n0.7*log(0.0072)/log(0.5)=t\n\nnow reach for the calculator: t=4.98 billion years.
271758
take logbase6 of 135 = x\n\nlog 135/log 6 = x\n2.737 = x\n\nThen substitute this into original to get:\n2^(log135/log6) + 1.3^(log135/log6)-2 \n\nIt should prove that it equals 30, which proves that both statements are ture.  However, I am getting that it is not equal to 30...so I think the proof is impossible,  unless your formula is typed incorrectly.