Create LogLogPlot of power function?

Questions
, ,
1

hello one and all,

I can not figure out why Mathematica will not plot this function under the LogLogPlot fxn. here is my code:

data = {{.1, 99}, {.02, 50.2}, {.004, 25.9}, {.0008, 13.4}, {.00016, 
   6.7}, {.000032, 4.1}, {.0000064, 3.1}}
p1 = ListLogLogPlot[data, PlotStyle -> {Red, PointSize[0.015]}];
p2 = ListPlot[data, PlotStyle -> {Red, PointSize[0.015]}];
q1 = Log[ems - a] == m Log[conc] + b
q1s = Solve[q1, ems];
q2 = Normal[q1s][[1, 1, 2]]
nlf1 = NonlinearModelFit[data, q2, {m, b, a}, conc];
nlf1P = nlf1["BestFitParameters"]
q3 = Normal[nlf1]
LogLogPlot[q3 /. nlf1P, {conc, 0, .11}]
Show[{p1, LogLogPlot[q3 /. nlf1P, {conc, 0, .11}]}]
Show[{Plot[q3, {conc, 0, .11}], p2}]

where the fit seems good and and plots well on the linear, linear plot, but not on the log, log plot. so let me know and thank you in advance, Lucas

2

Seems to work well for me:

ScreenShot2017-03-09at15.42.40.png
ScreenShot2017-03-09at15.42.40.png746Ă—1434 37.2 KB

Version of Mathematica, OS. Quit your kernel, and retry? If a,b,m, ems, or conc has a value the fit might not work.

3

Try Evaluate around

q3 /. nlf1P 
Evaluate[q3 /. nlf1P]

and

q3
Evaluate[q3]

in your plotting functions…

4

well, ok? I’m using Mathematica 11. I tried SetPrecision, that doesn’t do it. anything else, because I just get a blank.

5

ok, adding Evaluate worked. but why? wouldn’t any Plot function require an evaluation to plot the input function? Lucas

6

This behavior was introduced in 11.0.0 and was changed in 11.0.1 back to original. See also:

So if you update, it will be solved.

7

ok, thank you. you’re a big help. Lucas