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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: 20,747 Thanks: 2133 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: 1,310 Thanks: 551 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: 20,747 Thanks: 2133 If s, b, and a are all positive, s = ab^-t implies log(s) = log(a) - t*log(b). Tags converting, formula, linear, relationship Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post The Galin Algebra 1 February 21st, 2016 06:11 PM Karma Peny Math 28 November 13th, 2015 07:21 PM vyoung831 Linear Algebra 1 December 12th, 2014 08:20 PM JJP Linear Algebra 1 December 7th, 2014 04:21 PM Regnes Algebra 3 July 13th, 2011 04:56 PM

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