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 M₀, the mass after 12 hours is half that:
1/2 M₀ = M₀ exp(12k)
for some decay constant k. Solve for this k :
1/2 = exp(12k)
ln(1/2) = 12k
k = 1/12 ln(1/2) = - ln(2)/12
Now for some starting mass M₀, the mass M remaining after time t is given by
M = M₀ exp(kt )
So if M₀ = 590 g and t = 36 h, plugging these into the equation with the previously determined value of k gives
M = 590 exp(36k) = 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
