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.
mosses, fungi, molds and bacteria, and lichens
you guys are idiots if you dont know what they benefit from a rotin log so you tell me
After t hours the number of bacteria is 10*2t. So 10*2t > 3000000 => 2t > 300000 => t(log2) > log(300000) => t > log(300000)/(log2) => t > 18.19.. So in just over 18 hours (or 18 hours and 12 minutes, approx).
Insects, slime molds, worms, bacteria, protists, nematodes
Bacteria grow most rapidly during the log phase.
log(x) - log(6) = log(15)Add log(6) to each side:log(x) = log(15) + log(6) = log(15 times 6)x = 15 times 6x = 90
log(x6) = log(x) + log(6) = 0.7782*log(x) log(x6) = 6*log(x)
It really depends on the bacteria. Some multiply best at room temperature, some at 60 deg celcius. Bacteria has 4 phases in life. The lag, log/exponential, stationary or the death phase. Bacteria multiply best at its log phase. The log phase depends on the bacteria species.
log(f) + log(0.1) = 6 So log(f*0.1) = 6 so f*0.1 = 106 so f = 107
mosses, fungi, molds and bacteria, and lichens
log(x) + 4 - log(6) = 1 so log(x) + 4 + log(1/6) = 1 Take exponents to the base 10 and remember that 10log(x) = x: x * 104 * 1/6 = 10 x = 6/1000 or 0.006
you guys are idiots if you dont know what they benefit from a rotin log so you tell me
pH = -log[H+] = -log(3.7 x 10-6) = 5.43
You have, y = 6 + log x anti log of it, 10y = (106) x
bacteria
logx +7=1+log(x-1) 6=log(x-1)-logx 6=log[(x-1)/x] 10^6=(x-1)/x 1,000,000x=x-1 999,999x=-1 x=-1/999,999
-6