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So I have this exponential function that I don't really understand: LOGS

I have never studied anything to do with "r" and am attempting to use this information to find an average mass. I would have used a larger range of planets, but I could not find a larger data set that had what I would need, so instead of picking random planets to add I chose to keep with our solar system to remove further sampling bias.

Any help will be very appreciated

Alex Robinson
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  • It is an exponential function, but because your graph has a log scale, it looks like a line. Edit: However, it is unclear to me what the $x$-axis represents in your graph. Where did you get the graph from? Is there any more information about the $x$-axis that you have? – Zubin Mukerjee Aug 19 '17 at 23:20
  • @ZubinMukerjee I am aware of what happens when you put an exponential line of best fit on a log graph, my question is not why it is a line, but finding the mean of said line – Alex Robinson Aug 19 '17 at 23:21
  • I am also unsure what it means, because I don't know what the $x$-axis represents (see the edit to my original comment) – Zubin Mukerjee Aug 19 '17 at 23:26
  • just a nitpick $$10^{21}$$ kilograms is $$10^{24}$$ grams, aka a yottagram. see https://math.stackexchange.com/questions/834913/how-to-calculate-arithmetic-mean-of-log-values for help maybe ? –  Aug 19 '17 at 23:43
  • @ZubinMukerjee I made this graph, using Wikipedia for the information about the mass of those planets, how is the X axis confusing you? It's just planets ordered by mass – Alex Robinson Aug 20 '17 at 00:06
  • @Cursed1701 Usually, in graphs, the $y$ axis represents a quantity (in your graph, it does, and this quantity is mass), and the $x$ axis also represents a quantity (in your graph, it seems like the $x$ axis has no quantity associated with it) ... maybe for a more meaningful curve fit, you could have two variables that are quantities (like mass and distance from sun), instead of just one variable that is a quantity (in your case, mass). – Zubin Mukerjee Aug 20 '17 at 00:10
  • Also, what kind of average are you trying to find? If you want the arithmetic mean of the list of planet masses, then just add them up and divide by the number of planets in the list. – Zubin Mukerjee Aug 20 '17 at 00:13
  • @Cursed1701 the fact that the planets ordered by mass is very roughly exponentially distributed, probably doesn't mean much. I suspect that you could find a wide variety of other natural objects that have a very rough exponential distribution when ordered by mass, length, volume, etc. As an experiment, try taking the typical mass of mammals often found in zoos, sort them and plot – Χpẘ Aug 20 '17 at 00:14
  • @ZubinMukerjee ok, I understand your point and you do make a fair point. I am trying to use this data as a distribution of planet mass, not trying to show any correlation, but using this to then find the mean mass, assuming that any point along the line of best fit is equally likely, what does that make the mean mass? – Alex Robinson Aug 20 '17 at 00:17
  • @Cursed1701 Which type of mean do you want? The arithmetic mean? – Zubin Mukerjee Aug 20 '17 at 00:18
  • @ZubinMukerjee I don't know the terminology for different means. But the mean I want would be the sum of every (infinite) point on the line of best fit, and divided by that infinity, if that makes sense – Alex Robinson Aug 20 '17 at 00:21
  • To find the average value of a continuous function over some range $[a,b]$, you can take the integral of that function from $a$ to $b$, then divide by $b-a$. Edit: the process of "adding up infinite points and dividing by their number" is analogous to using an integral to find the average value of a function. – Zubin Mukerjee Aug 20 '17 at 00:24
  • The average value of a line is the midpoint of the line. So looking at this graph you have about 18000? However, in order to define average here, you need to define the area under the graph. And to do that, you need to give numerical values to the x-axis. You could choose to order the planets 1, 2, 3... But this choice is arbitrary, and thus your average is going to be arbitrary. In order to pull useful info out, you need more constraints. – Kaynex Aug 20 '17 at 13:25

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