# How to linearise y = ax b Demo for Swine Flu CW Using Logs to Linearise Curves.

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How to linearise y = ax b Demo for Swine Flu CW Using Logs to Linearise Curves

Linearising a power equation using logs y = ax b Swine Flu x246810121416182022 y9.524436487.9113141170200232265 This graph is NOT linear

Linearising a power equation using logs y = ax b y = ax b log y = log (ax b ) log y = log a + logx b log y = log a + blogx log y = blogx + log a So make a new table of values where Y = log y and X = log x This is of the form y = mx + c. Taking logs of both sides log(ab) = log(a) + log(b) log(a x ) = xlog(a) gradient = b y intercept = log a YmXc

x246810121416182022 y9.524436487.9113141170200232265 x=log x0.300.600.780.901.001.081.151.201.261.301.34 y=log y0.981.381.631.811.942.052.152.232.302.372.42 From the graph gradient y intercept m = 1.3962 c = 0.5485

gradient = m = 1.3962 = b y intercept = c = 0.5485 = log a y = ax b log y = blogx + log a Y = 1.3962X + 0.5485 a = 10 0.5485 = 3.54 0.5485  10 it = a Forwards and backwards a  log it  0.5485 YmXc

Using the equation If x = 5.5 find y y = 3.54×5.5 1.3962 = 38.2 Check if the answer is consistent with the table x = 5.5 find y y = 38.2 which is consistent with the table y = 3.54x 1.3962 x246810121416182022 y9.524436487.9113141170200232265

Using the equation If y = 100 find x 100 = 3.54x 1.3962 log100 = log(3.54x 1.3962 ) = log(3.54)+log(x 1.3962 ) = log(3.54)+1.3962log(x) y = 3.54x 1.3962 log(ab) = log(a) + log(b) log(a x ) = xlog(a) log both sides

log 100= log(3.54)+1.3962log(x) Forwards and backwards x  log it  ×1.3962  +log3.54 = log 100 log100  –log 3.54  ÷1.3962  10 it = x x = 10.97 x = 10.97 which is consistent with the table Check if the answer is consistent with the table y = 100 find x x246810121416182022 y9.524436487.9113141170200232265

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