Link to home
Start Free TrialLog in
Avatar of kacor
kacorFlag for Hungary

asked on

Formula of a line in log-log coordinate system

I have a line in log-log coordinate system that means all of it's points are known. In linear system I can simply write the formula of a line. Would be it possible in this case too?

Thanks for your help!

wbr
Janos
SOLUTION
Avatar of d-glitch
d-glitch
Flag of United States of America image

Link to home
membership
This solution is only available to members.
To access this solution, you must be a member of Experts Exchange.
Start Free Trial
ASKER CERTIFIED SOLUTION
Link to home
membership
This solution is only available to members.
To access this solution, you must be a member of Experts Exchange.
Start Free Trial
Avatar of wytcom
wytcom

My (later) post agrees with d-glitch.  Good fast job there, d-glitch!
Thanks wytcom.  I just wish I could spell  "eqatuion"  under pressure.

>> kacor

     It may not be immediately obvious, but function that generates a straight line on a log/log plot can only have one term.

     One way to see this is to realize that you only have two degrees of freedom (slope and intercept) and they wind up being gain and exponent.

      Another way is to realize that if there are more than one term, you won't get a straight line.
      The highest order term will dominate for large X, and the lowest order term will dominate for small X.

              For example:         Y  =  X^2 + B    has slope of 2 for X>>B, and a slope of 0 for X<<B

     
Avatar of kacor

ASKER

d-glitch and wytcom,

thanks for your help.

wbr
Janos
Avatar of kacor

ASKER

wytcom wrote in {http:#12413117}:

log(y) = m*log(x) + b

this is not true I mean because b is not value of a number but a length of log(b) (eg. distance, measured on logaritmic scale)

what do you think about?

Janos
Working forward this time:

               Y  =  Ao * X^m       This is a simple power law.


To plot this on log-log paper you have to start by taking the log of each side:

      log( Y)  =  log( Ao * X^m)    

                 =  log ( Ao)  +  log( X^m)             Use the properties of logs

                 =  log ( Ao)  +  m * log( X)             "         "           "         "

                 =  m * log( X)  +  log ( Ao)            Rearranage to see that this is in the form of  y = m*x + b  ==>  which is a straight line.
Avatar of kacor

ASKER

that's right but my question is which value have I to use here: b or log(b)
Avatar of kacor

ASKER

and some more:

If you know 2 points of the line you can calculate the slope:

m = (log(y2)-log(y1))/(log(x2)-log(x1))  
from this comes
m = log (y2/y1)/log(x2/x1))

but I don't know really that this m rests the same or I have to transform it too

and the last question is the question of either b or log(b) because:

if it is given the slope and the point P1(x1;y1) then b is calculable as follows

m*log(x) + b = 0 ->> m=-b/log(x)
or
m*log(x) + log(b) = 0 ->> m=-log(b)/log(x)  

which have I to choose??
If you have two points on a log-log plot, you should draw a line through them
and continue the line through the point  [ log(x)=0, log(Ao)] ==> [ x=1, Ao]

==================================================

Given two points, you can find m:

(log(y2)-log(y1))/(log(x2)-log(x1))   = m
----------------------------------------------------

Then you can find the gain term, since the line must also go through the point [ x=1, Ao]

(log(Ao)-log(y1))/(log(1)-log(x1))    = m
                                                         

log(Ao)  = m*[(log(1)-log(x1)] + log(y1)

             =  m*log(1/x1) + log(y1)


                       y1
     Ao    =   ---------
                    x1^m
b = log(y) as read off your log-log graph at the point where the line intersects with the vertical axis
This is correct:

m = (log(y2)-log(y1))/(log(x2)-log(x1))  
from this comes
m = log (y2/y1)/log(x2/x1))