Double Logarithmic Scale: Find/Create Equation

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To create an equation for a straight line on a double logarithmic scale graph, the format is y = m·log(log x) + b for a double-log graph and log y = m·log x + b for a log-log graph. To generate this equation, plot log x against log y, ensuring that the coordinates cover at least one magnitude to obtain a linear representation. The value of b can be determined by the y-intercept of the plotted line. If the result does not appear linear, reconsider the suitability of using log-log paper for the data. Understanding these principles is crucial for accurately interpreting relationships in logarithmic data.
Snow178
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double log scale - please help

I am trying to find/create the equation for a straight line on a double logarithmic scale graph - what is the format of the equation or how do I do this using data points from the graph.
 
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If you mean a log-log graph, a straight line would be \log y=m\cdot\log x + b. If you mean a double-log graph, a straight line would be y = m\cdot\log\log x + b. Simplify these as you see fit.
 
Yes - it is for the equation log y= m*logx+b - how do I generate this equation from a straight line on a log-log graph - particularly not sure about b.
 
If the paper is log log, then what you really want is the "y" values for the vertical axis, and the "x" values for the horizontal axis. Each of your coordinates should range over at least one magnitude; otherwise, you probably will not obtain a line. If the result is not a line, then the use of log log paper was probably not the correct choice (but then this is what you are trying to find out). Note carefully, you are plotting the x and y values here on log log paper - NOT their logarithms.
 
Snow178 said:
Yes - it is for the equation log y= m*logx+b - how do I generate this equation from a straight line on a log-log graph - particularly not sure about b.

Well, you plot logx against logy, in the same way that you would plot x against y. If you have an equation for a straight line y=mx+b, what would b be?
 
Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...

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