Timeline for How to correctly transform a log-log graph into untransformed exponential graph?
Current License: CC BY-SA 4.0
9 events
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| May 5, 2021 at 23:56 | vote | accept | slicer | ||
| May 5, 2021 at 0:43 | answer | added | Gregg H | timeline score: 1 | |
| May 4, 2021 at 2:37 | history | edited | kjetil b halvorsen♦ | CC BY-SA 4.0 |
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| May 3, 2021 at 18:08 | comment | added | whuber♦ | Your knowledge of logs and exponentials is incorrect. In particular, $\exp(1.19-0.116\log(x))=\exp(1.19)x^{0.116}.$ | |
| May 3, 2021 at 17:07 | comment | added | slicer | 1. I used natural log 2. I exponentiated on both sides of the linear log-log equation. So exp^(1.19 - 0.116 log(x))) which is exp(1.19) - exp(0.116 log(x)) which is 3.18 - 1.12 x ---> which is a linear line?? Am I doing something extremely stupid? | |
| May 3, 2021 at 17:02 | history | edited | slicer | CC BY-SA 4.0 |
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| May 3, 2021 at 15:59 | comment | added | Harvey Motulsky | 1. Please clarify log10 or natural log, to make sure the antilog transform is correct. 2. How did you transform the line to generate the curve that you said didn't work. The linear fit on the log log axes will be a curve on the back-transformed axes. | |
| May 3, 2021 at 15:03 | history | edited | slicer | CC BY-SA 4.0 |
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| May 3, 2021 at 14:57 | history | asked | slicer | CC BY-SA 4.0 |