guyz naveed1234 mentioned dat we had to multiply the gradient wid 10^5....i was abt to do the same thing in my paper but i think it is rong.....
U see....the table was smthng like this (i am not sure of the value but they were around the ones i rote)
(10^-5) ... lg d(10^-5) And to calculate gradient we had to subtract two lg d values(of course) and this value was the
31 1.49 denominator....any way the point is 1.49 - 1.15 = 0.34.............There is no need to do anything with
. 10^5.... i tell u why. if u use 31 X 10^-5 and use this full value to calculate its log, then it is -3.51 and
. similarly for for 14 X 10^-5 we get its log as -3.85. If we subtract the two...-3.51-(-3.85) we get the
14 1.15 same answer which is 0.34........
So dats why i THINK dat there is no need to multiply the gradient value wid anything....cuz we'll get the same answer either we use full value or the other. Anyway my gradient value was 1.27....and the constant p was smthng around 2.4 (and its postive) to get the value of p we had to do the inverse log of y-intercept......Neway.now next is Pure Mathemaics...