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Helppp Astro (1 Viewer)

SxC

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The star canopus has an aparent magnitude of -0.72 and an absolute magnitude of -3.1

(a)How far away is Canopus in parsecs?

m-M = 5log(d/10)

-0.72-(-3.1) = 5log(d/10)

5log(d/10) = 2.38

then it says (d/10) = 2.99 how did they get that???? i know its a stupid question but can anyone help me urgentlyyy plzzz
 

wind

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The log in that equation is to the base 10.

So to cancel that you use the notation 10^x rather than e^x.

Is the answer:
d = 29.923 pc?
 

pc_wizz

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use base 10 ... in the casio caculator use the 'log' button, not the 'ln' button
 

helper

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og(d/10)=0.476
d/10=inv log (0.476)


To inverse log, hit the inverse key then the log key. The answer should be 2.99

Then times by 10. Answer = 29.9 parsec
 

wind

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[shift] --> [log]

You'll get a notation looking like 10^x.

Just 10^x both sides of the equation and make "d" the subject to find the distance...simple maths :)
 

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