log(0) is not defined, so the first part of the question cannot be answered.log(5) = 0.6990 and log(1) = 0 so the reduction is 0.6990log(0) is not defined, so the first part of the question cannot be answered.log(5) = 0.6990 and log(1) = 0 so the reduction is 0.6990log(0) is not defined, so the first part of the question cannot be answered.log(5) = 0.6990 and log(1) = 0 so the reduction is 0.6990log(0) is not defined, so the first part of the question cannot be answered.log(5) = 0.6990 and log(1) = 0 so the reduction is 0.6990
noyhing but test - positive control
4-2.5=1.5 x 10^5
log(5)125 = log(5) 5^(3) = 3log(5) 5 = 3 (1) = 3 Remember for any log base if the coefficient is the same as the base then the answer is '1' Hence log(10)10 = 1 log(a) a = 1 et.seq., You can convert the log base '5' , to log base '10' for ease of the calculator. Log(5)125 = log(10)125/log(10)5 Hence log(5)125 = log(10) 5^(3) / log(10)5 => log(5)125 = 3log(10)5 / log(10)5 Cancel down by 'log(10)5'. Hence log(5)125 = 3 NB one of the factors of 'log' is log(a) a^(n) The index number of 'n' can be moved to be a coefficient of the 'log'. Hence log(a) a^(n) = n*log(a)a Hope that helps!!!!!
[log2 (x - 3)](log2 5) = 2log2 10 log2 (x - 3) = 2log2 10/log2 5 log2 (x - 3) = 2(log 10/log 2)/(log5/log 2) log2 (x - 3) = 2(log 10/log 5) log2 (x - 3) = 2(1/log 5) log2 (x - 3) = 2/log 5 x - 3 = 22/log x = 3 + 22/log 5
0.5
log(0) is not defined, so the first part of the question cannot be answered.log(5) = 0.6990 and log(1) = 0 so the reduction is 0.6990log(0) is not defined, so the first part of the question cannot be answered.log(5) = 0.6990 and log(1) = 0 so the reduction is 0.6990log(0) is not defined, so the first part of the question cannot be answered.log(5) = 0.6990 and log(1) = 0 so the reduction is 0.6990log(0) is not defined, so the first part of the question cannot be answered.log(5) = 0.6990 and log(1) = 0 so the reduction is 0.6990
Microbial load (cfu/g or cfu/ml) can be expressed as log10. So, if you have 100,000 microbes that is 5 log, 10,000 microbes is 4 log, 1,000 is 3 log, 100 microbes is 2 log and 10 microbes is 1 log. Now, if you went from 100,000 microbes cfu/g to 10,000 microbes cfu/g that would be a 1 log reduction (5 - 4 log). If you went from 100,000 to 32,000 that would be a 0.5 log reduction (5 - 4.5 log) and so on. I hope this helps St John Hall
noyhing but test - positive control
In studying the process of the sterility assurance level, which is how you guarantee the sterility of a sample of bacterium, each log reduction assures that 90% of the sample is sterile, so log 6 reduction just ensures it further.
4-2.5=1.5 x 10^5
No. log 20 is a positive number , so it you subtract it from log 5 you get less than log 5. However, log10 5 = 1 - log102 = 2- log1020 . or log 5 - log 20 = log 5 - log 4*5 = log 5 - (log 5 + log 4) = log 5 - log 5 - log 4 = - log 4 But we do not need to do all of these computations, because log 5 is different from log 5 - log 20 by the law of the equality that says two equals remain equal if and only if we subtract (in our case) the same thing from them.
7x = 5x log(7) = log(5)x = log(5) / (log(7) = 0.82709 (rounded)
[log2 (x - 3)](log2 5) = 2log2 10 log2 (x - 3) = 2log2 10/log2 5 log2 (x - 3) = 2(log 10/log 2)/(log5/log 2) log2 (x - 3) = 2(log 10/log 5) log2 (x - 3) = 2(1/log 5) log2 (x - 3) = 2/log 5 x - 3 = 22/log x = 3 + 22/log 5
100000 log = 5.
You can get a log of chew ( 5 cans ) for 15 dollars
log (6x + 5) = 26x + 5 > 06x + 5 - 5 > 0 - 56x > - 56x/6 > -5/6x > -5/6log (6x + 5) = 210^2 = 6x + 5100 = 6x + 5100 - 5 = 6x + 5 - 595 = 6x95/6 = 6x/695/6 = xCheck: