kacor
asked on
Formula of a line in log-log coordinate system going through P1 and P2
Hi Experts,
earlier I asked a similar question see
https://www.experts-exchange.com/questions/21183266/Formula-of-a-line-in-log-log-coordinate-system.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
https://www.experts-exchange.com/questions/21183266/Formula-of-a-line-in-log-log-coordinate-system.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
ASKER
thanks d-glitch I'll test it.
wbr Janos
wbr Janos
If you believe
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))
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))
ASKER CERTIFIED SOLUTION
membership
This solution is only available to members.
To access this solution, you must be a member of Experts Exchange.
ASKER
thanks that's right
In a normal plot, the equation for a straight line is:
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
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
In a normal plot, the equation for a straight line is:
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
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
ASKER
HI,
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
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
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