The given equations are incomprehensible, I'm afraid...
You're given that osmium-183 has a half-life of 12 hours, so for some initial mass <em>M</em>₀, the mass after 12 hours is half that:
1/2 <em>M</em>₀ = <em>M</em>₀ exp(12<em>k</em>)
for some decay constant <em>k</em>. Solve for this <em>k</em> :
1/2 = exp(12<em>k</em>)
ln(1/2) = 12<em>k</em>
<em>k</em> = 1/12 ln(1/2) = - ln(2)/12
Now for some starting mass <em>M</em>₀, the mass <em>M</em> remaining after time <em>t</em> is given by
<em>M</em> = <em>M</em>₀ exp(<em>kt </em>)
So if <em>M</em>₀ = 590 g and <em>t</em> = 36 h, plugging these into the equation with the previously determined value of <em>k</em> gives
<em>M</em> = 590 exp(36<em>k</em>) = 73.75
so 73.75 ≈ 74 g of Os-183 are left.
Alternatively, notice that the given time period of 36 hours is simply 3 times the half-life of 12 hours, so 1/2³ = 1/8 of the starting amount of Os-183 is left:
590/8 = 73.75 ≈ 74
