[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.
5 to the power of 200, written as 5^200, is a very large number. It is equal to 5 multiplied by itself 200 times. The result is a 401-digit number with a value of 7888609052210118054117285652827862296732064351090230047702789306640625.
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
with something called logarithms. So 1 = (1 + x)^5 log 1 = log ((1+x)^5) log 1 = 5 x log (1 +x) but log 1 = 0 therefore 0 = 5 x log(1+x) divide both sides by 5 and you get 0 = log (1+x) we know that log 1 = 0, therefore 1+ x = 1 and so x = 0
1.268293446
5.00
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.
The base of log, if unspecified, is taken to be 10 so you would be finding the value of the logarithm of 5 to the base 10.This is the value x, such that 10^x = 5.
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
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.
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:
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
5 to the power of 200, written as 5^200, is a very large number. It is equal to 5 multiplied by itself 200 times. The result is a 401-digit number with a value of 7888609052210118054117285652827862296732064351090230047702789306640625.
It is not possible.