Converting formula to linear relationship

Jan 2017
18
0
Britain
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 a moderator:

skipjack

Forum Staff
Dec 2006
21,481
2,470
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).
 
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May 2016
1,310
551
USA
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.
 
Jan 2017
18
0
Britain
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).
 
Jan 2017
18
0
Britain
Keep making errors

:furious:

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:eek:
 

skipjack

Forum Staff
Dec 2006
21,481
2,470
If s, b, and a are all positive, s = ab^-t implies log(s) = log(a) - t*log(b).