# 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?

wbr
Janos
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Commented:
Your log-log graph shows log(y) vs log(x).

Let s = log(y)
Let t = log(x)

For a line in the log-log graph we have
s = m*t + b

where m is the slope and b is the intecept with the vertical axis.

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

Or
y = exp( m* log(x) + b )
= exp(b)*x^m
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Commented:
The eqatuion of the line has to be some thing like

Y = Ao * X^ n

The slope of the line is n.

Once you know n, you can find Ao by inspection.
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Commented:
My (later) post agrees with d-glitch.  Good fast job there, d-glitch!
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Commented:
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

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retiredAuthor Commented:
d-glitch and wytcom,

wbr
Janos
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retiredAuthor Commented:
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)

Janos
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Commented:
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.
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retiredAuthor Commented:
that's right but my question is which value have I to use here: b or log(b)
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retiredAuthor Commented:
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??
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Commented:
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
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Commented:
b = log(y) as read off your log-log graph at the point where the line intersects with the vertical axis
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Commented:
This is correct:

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