Question 548456
I assume you meant {{{M=6-(5/2)log(I/I[0])}}}
For a star of magnitude 1,
{{{1=6-(5/2)log(I/I[0])}}} --> {{{1-6=-(5/2)log(I/I[0])}}} --> {{{-5=-(5/2)log(I/I[0])}}}
Sorry, if I'm going to slow. I'd rather err on the side of over-explaining.
Then {{{5=(5/2)log(I/I[0])}}} --> {{{1=(1/2)log(I/I[0])}}} --> {{{2=log(I/I[0])}}} --> {{{10^2=100=I/I[0]}}} --> {{{I=100*I[0]}}}
For a star of magnitude 6,
{{{6=6-(5/2)log(I/I[0])}}} --> {{{6-6=-(5/2)log(I/I[0])}}} --> {{{0=-(5/2)log(I/I[0])}}}
So {{{0=log(I/I[0])}}} --> {{{10^0=1=I/I[0]}}} --> {{{I=I[0]}}}
So the light from a star of magnitude 1 is 100 times more intense than the light from a star of magnitude 6.