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 August 11th, 2017, 05:34 AM #1 Newbie   Joined: Jan 2017 From: Britain Posts: 18 Thanks: 0 Converting formula to linear relationship I am struggling with converting formulae to the linear form Y = mX +C For example if I have the relationship ay=b^x and I need to express this in linear form using logs, I get: logy+ loga = xlogb not sure how to get this to the form Y=mX+C because if I move the loga onto the other side I get: logY = xlogb - loga which would correspond to -C Help appreciated. Thanks. Last edited by skipjack; August 11th, 2017 at 06:15 AM.
 August 11th, 2017, 06:20 AM #2 Global Moderator   Joined: Dec 2006 Posts: 19,704 Thanks: 1804 The "- log(a)" is equivalent to "+ (-log(a))", so C = -log(a). Of course, your Y is log(y) and your m is log(b). Thanks from JeffM1
 August 11th, 2017, 06:22 AM #3 Senior Member   Joined: May 2016 From: USA Posts: 1,148 Thanks: 479 I do not understand the question. It is false that $ay = b^x \implies y = mx + c.$ The relationship between y and x is exponential rather than linear. Furthermore in the general linear form of y = mx + c, c can be a negative number. Adding a negative number is equivalent to subtracting a positive number. (And m may be a negative number as well.) Let's do this in steps. $ay = b^x \implies log(ay) = log(b^x) \implies$ $log(a) + log(y) = xlog(b) \implies log(y) = log(b) * x - log(a) \implies$ $u = mx + c, \text { where } u = log(y),\ m = log(b),\ \text { and } c = -\ log(a).$ Clear now? EDIT: Very long winded way to say what skipjack already said.
 August 11th, 2017, 06:40 AM #4 Newbie   Joined: Jan 2017 From: Britain Posts: 18 Thanks: 0 Thanks to both of you. I think I phrased the question badly -I meant that it was the 'reduction' of an exponential relationship to a linear one. Thanks for the clear explanation-I was thinking it had to be +C always! ( of course i knew m could be -m as a negative gradient).
 August 12th, 2017, 08:58 AM #5 Newbie   Joined: Jan 2017 From: Britain Posts: 18 Thanks: 0 Keep making errors Not sure what I'm doing wrong in this reduction of formula to linear law: The formula given with data is: s = ab^-t t: 1 2 3 4 s: 1.5 0.4 0.1 0.02 What I have done is tried to convert the formula like this: logS = -tlogb = loga with logb as 'm', loga as 'c' so I then plotted -t against logS: -t: -1 -2 -3 -4 log's': 0.2 -0.4 -1 -1.7 The answer in the book is a=6 and b = 4 so the gradient is 4 but m= (taking sets of points) -1.7 -(-1)/-4-(-3) = -0.7/-1 = 0.7!!!! I'm clearly getting things badly wrong here!!! Help please
 August 12th, 2017, 02:33 PM #6 Global Moderator   Joined: Dec 2006 Posts: 19,704 Thanks: 1804 If s, b, and a are all positive, s = ab^-t implies log(s) = log(a) - t*log(b).

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