SOLUTION: This is one of the most confusing problems I have ever read. Please help me. The apparent magnitude, M of a celestial body can be approximated by the formula M=6-2.5 log I/I Whe

Question 450140: This is one of the most confusing problems I have ever read. Please help me.
The apparent magnitude, M of a celestial body can be approximated by the formula
M=6-2.5 log I/I
Where I is the intensity of light as measured from earth, and I0 is the approximate intensity light of light from the faintest star, visible by the naked eye. Find M if I and I0=1
What apparent magnitude of a star whose light as measured from earth is 1,000,000 times as intense as the faintest stars visible to the naked eye.
I'd appreciate any help. I don't seem to be able to understand this problem verbally or algebraically.
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
The apparent magnitude, M of a celestial body can be approximated by the formula
M=6-2.5 log I/Io
========================================================================
Note: I could not tell from your post if the fraction was I/Io or Io/I
========================================================================
Where I is the intensity of light as measured from earth, and Io is the approximate intensity light of light from the faintest star, visible by the naked eye. Find M if I and Io=1
---
M = 6-2.5*log[1/1]
M = 6-2.5*0
M = 6
---------------------------
What apparent magnitude of a star whose light as measured from earth is 1,000,000 times as intense as the faintest stars visible to the naked eye.
----
M = 6-2.5log[1,000,000Io/Io]
---
M = 6-2.5*log[10^6]
---
M = 6-2.5*6 = 6-15 = -9
----
Note: If the fraction is really Io/I the answer would be
as follows:
M = 6-2.5*log[10^-6]
= 6-2.5*(-6) = 6+15 = 21
==========================
Cheers,
Stan H.
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