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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

Thanks for your help!

wbr

Janos

Y = Ao * X^ n

The slope of the line is n.

Once you know n, you can find Ao by inspection.

>> 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

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

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.

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

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

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??

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)

--------------------------

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

= m*log(1/x1) + log(y1)

y1

Ao = ---------

x1^m

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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