450 



For the point A when J- = 1-297, -||^ - 0-015 

 then' a +0-0156= 1-297 



For the point B when -^- = 2-037, 



x 



then a + 0-0556 = 2-037 



Subtracting (1) from (2) 0-04& = 0-738 



b = 18-45 

 and a = 1-297 - 0-015 x 18-45 



= 1-020 



The law is y = 1-020 2 + 18-45 Iog 10 a:. 



(b) Dividing throughout by log 10 tT ; the law becomes 



y 



_ i 



~ 



Hence, plotting 



vertically and 



CC 



Iog 10 n? 

 a straight line if the values follow the law 



horizontally will give 



y 



Iog 10 a; 



17-93 



44-83 

 25-57 



52-50 

 32-48 



62-91 



42-87 



73-65 



53-74 



87-04 



67-79 



99-56 



80-13 



Fig. 147 shows the resulting straight line. 



Let A and B be two points taken on this straight line. 



= 38-8, 



For the point A when 



then 

 For the point B when 



then b + 80a = 100-0 



Iog 10 a ' Iog 10 



I + 20a = 38-8 . , 



11 .i"2 



_ 20 



= 100-0, 



= 80, 



(1) 



(2) 



Subtracting (1) from (2) 60a = 61-2 



a - 1-02 

 and b = 38-8 - 20 x 1-02 



= 18-40 

 The law is y = 1-02 2 + 18-40 Iog 10 0. 



