Equation of Straight Line on the Log-Log ScaleDate: 03/06/2006 at 00:41:55 From: hard stone Subject: straight line equation on the log-log scale I have a log-log graph with a straight line on it, and I want to find the line's equation. The x-axis is scaled as 0.01, 0.1, 1, 10, 100 and the y-axis is 10, 100, 1000, 10000. I am using two points on the given line, (100,1) and (600,0.2) and finding the slope m = -1 and the y-intercept b = 101 so my equation is y = -1x + 101. But none of the points on the given line check when I put them in that equation except (100,1), so that can't be the right equation. What am I doing wrong? Date: 03/06/2006 at 09:51:17 From: Doctor Peterson Subject: Re: straight line equation on the log-log scale Hi, Hard Stone. A straight line on the log-log scale is not given by y = mx + b but by log(y) = m log(x) + b since the actual locations on the graph are the logs of the coordinates x and y. This in turn means the same as y = 10^b x^m (by raising 10 to the power on each side). So in order to find the parameters of this equation, which I'll simplify to y = c x^m you want to plug x and y for two points into either form of the equation and solve for the parameters. In this case, you can solve 1 = c 100^m 0.2 = c 600^m You might start by dividing one equation by the other to eliminate c, and then take a log. Alternatively, you can solve log(1) = m log(100) + b log(0.2) = m log(600) + b which might be more familiar. If you have any further questions, feel free to write back. - Doctor Peterson, The Math Forum http://mathforum.org/dr.math/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994-2013 The Math Forum
http://mathforum.org/dr.math/