Pharmacology for Anaesthesia and Intensive Care

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Section IBasic principles

understand the different pharmacokinetic models available to choose one that suits

the individual patient and their surgical intervention.

Appendix

Remember:

w×z= 10 x× 10 y= 10 (x+y)

w÷z= 10 x÷ 10 y= 10 (x−y).


Forexample, multiply 13 by 257 using their logarithms:

[NB log(13)= 1 .1139; i.e. 13 = 101.^1139 and log(257)= 2 .4100; i.e. 257 = 102.^4100 ]

13 × 257 = 101.^1139 × 102.^4100

= 10 (1.^1139 +^2 .4100)

= 103.^5239.

Toconvert back to a numerical value we need to find the antilog of 3.5239 in the

antilog table. However, only the numbers 0.0001 to 0.9999 are contained within

antilog tables so we have to split the exponent into an integer part and a positive

decimal part; for the example we used this gives 10^3 × 10 0.5239.Weknow this is 1000

×antilog(0.5239); we can look up 0.5239 in the body of the logarithm tables and find

it corresponds to 3.341.

Sothe result of multiplying 13 by 257 is 1000×3.341, which is 3341.

Forexample, divide 13 by 257 using their logarithms:

[NB log(13)= 1 .1139; i.e. 13 = 101.^1139 and log(257)= 2 .4100; i.e. 257 = 102.^4100 ]

13 ÷ 257 = 101.^1139 ÷ 102.^4100

= 10 (1.^1139 −^2 .4100)

= 10 (−^1 .2961).

Although this can be written 10−^1 × 10 −0.2961,which is 0.1×antilog(−0.2961), there

are no tables of negative logarithms!

Therefore, we actually write it as 10−^2 × 10 0.7039because− 2 +0.7039=−1.2961.

Wenow have 0.01×antilog(0.7039) and the antilog of 0.7039 is 5.057, so the result

of dividing 13 by 257 is 0.01×5.057, which is 0.05057.

Of course, this is a very old-fashioned way of finding logs and antilogs – a calculator

can do both with much greater accuracy and without the need to change negative

values to the sum of an integer part and a positive decimal part as we did in the

second calculation. In fact, these days we would never dream of using logarithms for

such calculations – that is what a calculator is for! However, it has introduced us to

the idea of logarithms and shown that the logarithm to the base 10 of a number is

the same as its exponent when it is written as a power of 10.