Logarithms APPENDIX IV. 265 Appendix

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1 APPENDIX IV Logarithms Sometimes, a umerical expressio may ivolve multiplicatio, divisio or ratioal powers of large umbers. For such calculatios, logarithms are very useful. They help us i makig difficult calculatios easy. I Chemistry, logarithm values are required i solvig problems of chemical kietics, thermodyamics, electrochemistry, etc. We shall first itroduce this cocept, ad discuss the laws, which will have to be followed i workig with logarithms, ad the apply this techique to a umber of problems to show how it makes difficult calculatios simple. We kow that = 8, = 9, = 1, 0 = 1 I geeral, for a positive real umber a, ad a ratioal umber m, let a m = b, where b is a real umber. I other words the m th power of base a is b. Aother way of statig the same fact is logarithm of b to base a is m. If for a positive real umber a, a 1 a m = b, we say that m is the logarithm of b to the base a. We write this as b log a = m, log beig the abbreviatio of the word logarithm. Thus, we have log 8 =, Sice = 8 log 9 =, Sice = 9 1 log =, Sice = 1 log 1 = 0, 0 Sice = 1 Laws of Logarithms I the followig discussio, we shall take logarithms to ay base a, (a > 0 ad a 1) First Law: log a (m) = log a m + log a Proof: Suppose that log a m = x ad log a = y The a x = m, a y = Hece m = a x.a y = a x+y It ow follows from the defiitio of logarithms that log a (m) = x + y = log a m log a m Secod Law: log a = log a m log a Proof: Let log a m = x, log a = y 6 Appedix

2 The a x = m, a y = m Hece Therefore x a = = a y a x y m loga = x y = loga m loga Third Law : log a (m ) = log a m Proof : As before, if log a m = x, the a x x x m = a = a The ( ) = m givig log a (m ) = x = log a m Thus accordig to First Law: the log of the product of two umbers is equal to the sum of their logs. Similarly, the Secod Law says: the log of the ratio of two umbers is the differece of their logs. Thus, the use of these laws coverts a problem of multiplicatio / divisio ito a problem of additio/ subtractio, which are far easier to perform tha multiplicatio/divisio. That is why logarithms are so useful i all umerical computatios. Logarithms to Base 10 Because umber 10 is the base of writig umbers, it is very coveiet to use logarithms to the base 10. Some examples are: log = 1, sice 10 1 = 10 log =, sice 10 = 100 log = 4, sice 10 4 = log =, sice 10 = 0.01 log =, sice 10 = ad log 10 1 = 0 sice 10 0 = 1 The above results idicate that if is a itegral power of 10, i.e., 1 followed by several zeros or 1 preceded by several zeros immediately to the right of the decimal poit, the log ca be easily foud. If is ot a itegral power of 10, the it is ot easy to calculate log. But mathematicias have made tables from which we ca read off approximate value of the logarithm of ay positive umber betwee 1 ad 10. Ad these are sufficiet for us to calculate the logarithm of ay umber expressed i decimal form. For this purpose, we always express the give decimal as the product of a itegral power of 10 ad a umber betwee 1 ad 10. Stadard Form of Decimal We ca express ay umber i decimal form, as the product of (i) a itegral power of 10, ad (ii) a umber betwee 1 ad 10. Here are some examples: (i). lies betwee 10 ad 100. = = (ii) lies betwee 1000 ad = = (iii) 0.00 lies betwee ad = ( ) 10 =.0 10 (iv) lies betwee ad = ( ) 10 4 = Chemistry 66

3 I each case, we divide or multiply the decimal by a power of 10, to brig oe o-zero digit to the left of the decimal poit, ad do the reverse operatio by the same power of 10, idicated separately. Thus, ay positive decimal ca be writte i the form = m 10 p where p is a iteger (positive, zero or egative) ad 1< m < 10. This is called the stadard form of. Workig Rule 1. Move the decimal poit to the left, or to the right, as may be ecessary, to brig oe o-zero digit to the left of decimal poit.. (i) If you move p places to the left, multiply by 10 p. (ii) If you move p places to the right, multiply by 10 p. (iii) If you do ot move the decimal poit at all, multiply by (iv) Write the ew decimal obtaied by the power of 10 (of step ) to obtai the stadard form of the give decimal. Characteristic ad Matissa Cosider the stadard form of = m 10 p, where 1 < m < 10 Takig logarithms to the base 10 ad usig the laws of logarithms log = log m + log 10 p = log m + p log 10 = p + log m Here p is a iteger ad as 1 < m < 10, so 0 < log m < 1, i.e., m lies betwee 0 ad 1. Whe log has bee expressed as p + log m, where p is a iteger ad 0 log m < 1, we say that p is the characteristic of log ad that log m is the matissa of log. Note that characteristic is always a iteger positive, egative or zero, ad matissa is ever egative ad is always less tha 1. If we ca fid the characteristics ad the matissa of log, we have to just add them to get log. Thus to fid log, all we have to do is as follows: 1. Put i the stadard form, say = m 10 p, 1 < m <10. Read off the characteristic p of log from this expressio (expoet of 10).. Look up log m from tables, which is beig explaied below. 4. Write log = p + log m If the characteristic p of a umber is say, ad the matissa is.41, the we have log = +.41 which we ca write as.41. If, however, the characteristic p of a umber m is say ad the matissa is.41, the we have log m = We caot write this as.41. (Why?) I order to avoid this cofusio we write for ad thus we write log m =.41. Now let us explai how to use the table of logarithms to fid matissas. A table is appeded at the ed of this Appedix. Observe that i the table, every row starts with a two digit umber, 10, 11, 1,. 9, 98, 99. Every colum is headed by a oe-digit umber, 0, 1,,.9. O the right, we have the sectio called Mea differeces which has 9 colums headed by 1, Now suppose we wish to fid log (6.4). The look ito the row startig with 6. I this row, look 6 Appedix

4 at the umber i the colum headed by. The umber is 94. This meas that log (6.0) = 0.94* But we wat log (6.4). So our aswer will be a little more tha How much more? We look this up i the sectio o Mea differeces. Sice our fourth digit is 4, look uder the colum headed by 4 i the Mea differece sectio (i the row 6). We see the umber there. So add to 94. We get 948. So we fially have log (6.4) = Take aother example. To fid log (8.1), we look i the row 81 uder colum, ad we fid We cotiue i the same row ad see that the mea differece uder is 4. Addig this to 9096, ad we get So, log (8.1) = Fidig N whe log N is give We have so far discussed the procedure for fidig log whe a positive umber give. We ow tur to its coverse i.e., to fid whe log is give ad give a method for this purpose. If log = t, we sometimes say = atilog t. Therefore our task is give t, fid its atilog. For this, we use the readymade atilog tables. Suppose log =.. To fid, first take just the matissa of log. I this case it is.. (Make sure it is positive.) Now take up atilog of this umber i the atilog table which is to be used exactly like the log table. I the atilog table, the etry uder colum i the row. is 44 ad the mea differece for the last digit i that row is, so the table gives 44. Hece, atilog (.) =.44 Now sice log =., the characteristic of log is. So the stadard form of is give by = or = 44. Illustratio 1: If log x = 1.01, fid x. Solutio: We fid that the umber correspodig to 01 is 119. Sice characteristic of log x is 1, we have x = = 11.9 Illustratio : If log x =.1, fid x. Solutio: From atilog tables, we fid that the umber correspodig to 1 is 166. Sice the characteristic is i.e.,, so x = = Use of Logarithms i Numerical Calculatios Illustratio 1: Fid Solutio: Let x = The log x = log (6. 1.9) = log 6. + log 1.9 Now, log 6. = 0.99 log 1.9 = log x = , Takig atilog * It should, however, be oted that the values give i the table are ot exact. They are oly approximate values, although we use the sig of equality which may give the impressio that they are exact values. The same covetio will be followed i respect of atilogarithm of a umber. Chemistry 68

5 x = 8.1 Illustratio : 1. (1.) Fid 11.. Solutio: Let x = The log x = log (1.) 11.. (1.) 11.. = log 1. log (11..) = log 1. log 11.. Now, log 1. = log 1. = log 11. = log. = 1.11 log x = =.14 x = Illustratio : Fid (1.4) 6 (.) 1 Solutio: Let x = (1.4) 6 (.) 1 The log x = 1 log (1.4) 6 (.) 1 = 1 [log (1.4) + log 6 log (.) log 1] = log log 6 log. 1 log Now, usig log tables log 1.4 = 1.8 log 6 = 1.48 log. = 0.61 log 1 = 1. log x = log (1.8) + 1 (1.48) (0.61) 1 (1.) 4 4 =.4 x = Appedix