
Algebra PreAlgebra and Basic Algebra Math Forum 
 LinkBack  Thread Tools  Display Modes 
August 11th, 2017, 06:34 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 07:15 AM. 
August 11th, 2017, 07:20 AM  #2 
Global Moderator Joined: Dec 2006 Posts: 19,964 Thanks: 1849 
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). 
August 11th, 2017, 07:22 AM  #3 
Senior Member Joined: May 2016 From: USA Posts: 1,206 Thanks: 494 
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, 07:40 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 explanationI was thinking it had to be +C always! ( of course i knew m could be m as a negative gradient). 
August 12th, 2017, 09:58 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, 03:33 PM  #6 
Global Moderator Joined: Dec 2006 Posts: 19,964 Thanks: 1849 
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  
Display Modes  

Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
Can you express this relationship as a formula?  The Galin  Algebra  1  February 21st, 2016 07:11 PM 
The relationship between 1+2+3+4+... and 1/12  Karma Peny  Math  28  November 13th, 2015 08:21 PM 
Relationship between singular matrices and linear dependency?  vyoung831  Linear Algebra  1  December 12th, 2014 09:20 PM 
Need help with converting 2 nonlinear equations to linear  JJP  Linear Algebra  1  December 7th, 2014 05:21 PM 
Converting a linear graph to a linear equation?  Regnes  Algebra  3  July 13th, 2011 05:56 PM 