Log–log plot: Difference between revisions

Revision as of 17:23, 4 April 2021

Revision as of 13:06, 1 May 2021

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To find the slope of the plot, two points are selected on the ''x''-axis, say ''x''₁ and ''x''₂. Using the above equation:

:<math> \log[F (x_1)] = m \log (x_1) + b, \, </math>

and

:<math> \mathrm {log}[F (x_2)] = m \log (x_2) + b. \, </math>

The slope ''m'' is found taking the difference:

:<math> m = \frac { \mathrm {log} (F_2) - \mathrm {log} (F_1)} { \log(x_2) - \log(x_1) } = \frac {\log (F_2/F_1)}{\log(x_2/x_1)}, \,</math>

where ''F''₁ is shorthand for ''F'' ( ''x''₁ ) and ''F''₂ is shorthand for ''F'' ( ''x''₂ ). The figure at right illustrates the formula. Notice that the slope in the example of the figure is ''negative''. The formula also provides a negative slope, as can be seen from the following property of the logarithm:

:<math> \log(x_1/x_2) = -\log(x_2/x_1). \, </math>

=== Finding the function from the log–log plot ===

The above procedure now is reversed to find the form of the function ''F''(''x'') using its (assumed) known log–log plot. To find the function ''F'', pick some ''fixed point'' (''x''₀, ''F''₀), where ''F''₀ is shorthand for ''F''(''x''₀), somewhere on the straight line in the above graph, and further some other ''arbitrary point'' (''x''₁, ''F''₁) on the same graph. Then from the slope formula above:

::<math> m = \frac {\log (F_1 / F_0)}{\log(x_1 / x_0)} </math>

which leads to

::<math> \log(F_1 / F_0) = m \log(x_1 / x_0) = \log[(x_1 / x_0)^m ]. \, </math>

Notice that 10^{log₁₀(''F''₁)} = ''F''₁. Therefore, the logs can be inverted to find:

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or

: <math>F_1 = \frac{F_0}{x_0^m} \,\, x^m, \, </math>

which means that

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In other words, ''F'' is proportional to ''x'' to the power of the slope of the straight line of its log–log graph. Specifically, a straight line on a log–log plot containing points (''F''₀,&nbsp;''x''₀) and (''F''₁,&nbsp;''x''₁) will have the function:

:<math> F(x) = {F_0}\left(\frac{x}{x_0} \right)^\frac {\log (F_1/F_0)}{\log(x_1/x_0)}, </math>

Of course, the inverse is true too: any function of the form

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: <math> \mathrm{constant} = \frac{F_0}{x_0^m} </math>

Substituting back into the integral, you find that for A over ''x''₀ to ''x''₁

: <math> A = \frac{F_0/x_0^m}{m+1}\cdot (x_1^{m+1}-x_0^{m+1}) </math>

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Therefore: <math> A = \frac{F_0}{m+1} \cdot \left[x_1 \cdot \left(\frac {x_1}{x_0}\right)^m - x_0\right] </math>

For ''m''&nbsp;=&nbsp;−1, the integral becomes <math display="block"> A_{(m=-1)} = \int_{x_0}^{x_1} F(x) = \int_{x_0}^{x_1} \frac {\mathrm{constant}}{x} = \frac{F_0}{x_0^{-1}} \int_{x_0}^{x_1} \frac {1}{x} = F_0 \cdot x_0 \cdot \ln x: [x_0,x_1] </math>

: <math> A_{(m=-1)} = F_0 \cdot x_0 \cdot \ln \frac{x_1}{x_0}</math>