The equation of a straight line on a log-log plot must have the form:

Y = Ao * X^m

From the earlier question we have:

m = log (Y2/Y1)/log(X2/X1))

Y1 Y2

Ao = --------- = --------

X1^m X2^m

Solved

Posted on 2004-11-03

Hi Experts,

earlier I asked a similar question see

http://www.experts-exchange.com/Miscellaneous/Math_Science/Q_21183266.html#12444220

But now my question is: what is the formula of a line in log-log coordinate system going through P1(X1;Y1) and P2(X2;Y2) because I've tried to solve it but I was unsuccesful. Thanks for your help!

Janos

earlier I asked a similar question see

http://www.experts-exchang

But now my question is: what is the formula of a line in log-log coordinate system going through P1(X1;Y1) and P2(X2;Y2) because I've tried to solve it but I was unsuccesful. Thanks for your help!

Janos

8 Comments

The equation of a straight line on a log-log plot must have the form:

Y = Ao * X^m

From the earlier question we have:

m = log (Y2/Y1)/log(X2/X1))

Y1 Y2

Ao = --------- = --------

X1^m X2^m

Y = Ao * X^m

Then you have two equations in two unknowns:

Y1 = Ao * X1^m

Y2 = Ao * X2^m

You can solve either one or both to find Ao in terms of X,Y, and m

Y1 Y2

Ao = --------- = --------

X1^m X2^m

You can use the second equality (which has eliminated Ao) to solve for m in terms of X and Y.

Y1 Y2

--------- = --------

X1^m X2^m

You did this yourself in the earlier question:

m = log (Y2/Y1)/log(X2/X1))

Either one works to find a solution, or to gain insight.

But trying to combine them on the fly can be a little confusing.

y=a*x+b

In a log-log plot it will be:

logy=a*logx+b =>

y=10^(a*logx+b) =>

y=10^(a*logx)*10^b =>

y=(10^logx)^a*10^b => (because if y=logx <=> x=10^y=10^logx)

y=x^a*10^b

Thus, this function will give a linear graph in a log-log plot

y=a*x+b

where a=(Y2-Y1)/(X2-X1) (the slope) and b=(X1*Y2-X2*Y1)/(X1-X2) (intersection with y-axis)

In a log-log plot it will be:

logy=a*logx+b =>

y=10^(a*logx+b) =>

y=[10^(a*logx)]*(10^b) =>

y=[(10^logx)^a]*(10^b) => (because if y=logx <=> x=10^y=10^logx)

y=(x^a)*(10^b)

Thus, this function will give a linear graph in a log-log plot

I tried to get the needed equation but my trying was unsuccesful when I used the above said:

For example there are taken 2 points:

Point 1: y1=10; x1=4,1

Point 2: y2=0,1; x2=176

I tried to insert the values.

the slope is m = log (Y2/Y1)/log(X2/X1)) =>

m = log10(0,1/10) / log10(176/4) m = -2 / 1,632729 = -1,224943

the intersection point with y-axis is b = y1 / x1^m

b = 10 / 4,1^(-1,224943) = 10 / 0,177572 = 56,31517

and using the equation of y=(x^a)*(10^b) = 3,7E+55 I've got a bad value.

but using the equation of y = b * (x^a) the results are OK.

thanks for your support

Janos

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