It is the value for the affinity for a fatty or water phase. It is the ratio of the amount of compound in the octanol:water phase. It is also called LogPOW or LogPO:W.
They should have a log P value of 2.6.
Hold on to your hat! Suppose the rate of depreciation is p percent per year. Current value = Start Value*(1 - percentage/100)years That gives 2047.08 = 5500*(1 - p/100)9 So 0.3722 = (1 - p/100)9 or log(0.3722) = 9*log(1-p/100) -0.4292 = 9*log(1-p/100) -0.0477 = log(1-p/100) 10-0.0477 = 1-p/100 0.8960 = 1-p/100 p/100 = 1 - 0.8960 = 0.1040 and, finally, p = 0.1040*100 = 10.4%
Suppose you have a variable which grows at r% per period [year] and you want to find out how many periods, t, it will take before it grows, from a starting value of P to reach the value Q.The approximate method used by younger pupils is trial and error. This is because the y do not have the necessary mathematical knowledge.After t periods, the variable in question will have increased from P to Q whereQ = P*(1+r/100)tthen Q/P = (1+r/100)ttaking logarithms, log(Q/P) = log[(1+r/100)t] = t*log(1+r/100)and so finally, t = log(Q/P) / log(1+r/100).
Yes. Take any rational number p. Let a = any number that is not a power of 10, so that log(a) is irrational. and let b = p/log(a). log(a) is irrational so 1/log(a) must be irrational. That is, both log(a) and log(b) are irrational. But log(a)*log(b) = log(a)*[p/log(a)] = p which is rational. In the above case all logs are to base 10, but any other base can be used.
The value of log o is penis
The logarithm of the octanol/water partition coefficient (log P) of boswellia serrata has been reported to be around 7.5. The pKa value of boswellic acid, a major compound found in boswellia serrata, is around 4.8.
determination of log table value
If it is compounded annually, then: F = P*(1 + i)^t {F is final value, P is present value, and i is interest rate, t is time}.So if it triples, F/P = 3, and 12 years: t = 12, so we have 3 = (1 + i)^12, solve for i using logarithms (any base log will do, but I'll use base 10):log(3) = log((1+i)^12) = 12*log(1+i)(log(3))/12 = log(1+i).Now take 10 raised to both sides: 10^((log(3))/12) = 10^log(1+i) = 1 + ii = 10^((log(3))/12) - 1 = 0.095873So a rate of 9.5873 % (compounded annually) will triple the investment in 12 years.
log(21.4) = 1.330413773
log(22) = 1.342422681
Compounds with a log P value of 5 are highly lipophilic, meaning they have a strong tendency to partition into lipid environments over aqueous ones. This high log P indicates that such compounds may have low solubility in water, which can affect their bioavailability and absorption in biological systems. Additionally, their lipophilicity can influence their distribution, metabolism, and potential toxicity. Consequently, while they may exhibit desirable properties for certain applications, they also pose challenges in drug formulation and delivery.
log(0.99) = -0.004364805