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 Mar 23rd 2016, 06:58 PM #1 Member   Join Date: Mar 2016 Location: Manchester Posts: 37 Vaucouleurs Profile I am having a little trouble understanding some maths in a problem involving the surface brightness of galaxies. I know of the theory behind the problem but I don't know where exactly one step comes from in terms of the maths. In the attachment, from the first to the second line, it goes from -2.5log(e^-7.67[………etc.,) on the first line to 7.67log e x [(R\Re)^1\4 -1] on the second line. I was hoping maybe someone could show me this step done properly so that I can understand the mathematics. Attached Thumbnails
Mar 23rd 2016, 07:57 PM   #2

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 Originally Posted by Clyner82 I am having a little trouble understanding some maths in a problem involving the surface brightness of galaxies. I know of the theory behind the problem but I don't know where exactly one step comes from in terms of the maths. In the attachment, from the first to the second line, it goes from -2.5log(e^-7.67[………etc.,) on the first line to 7.67log e x [(R\Re)^1\4 -1] on the second line. I was hoping maybe someone could show me this step done properly so that I can understand the mathematics.
It uses the law: log(ab) = log(a) + log(b). Here we have a = I_e and b = e^(-7.76 ((R/R_e)^(1/4)-1)

-Dan
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 Mar 23rd 2016, 08:47 PM #3 Member   Join Date: Mar 2016 Location: Manchester Posts: 37 I am still not seeing how you get from step 1 to step 2
 Mar 24th 2016, 09:17 AM #4 Physics Team     Join Date: Jun 2010 Location: Morristown, NJ USA Posts: 2,320 C-2.5 log I_c is apparently a constant, so that becomes "const" in line 2. They also use the principle that log a^b = b log a. In this case b = that rather complicated expression, which for some reason they break in two and put part of it in front of the log term and some after. In essence it's this: log a^(bc) = b x c x (log a) = b x (log a) x c Last edited by ChipB; Mar 24th 2016 at 02:30 PM.
 Mar 24th 2016, 09:39 AM #5 Member   Join Date: Mar 2016 Location: Manchester Posts: 37 Thanks a millions Chip, that makes more sense now. Much appreciate your time in helping me out.

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