How do you express the concept of the rate of decay of a radioactive substance as a differential equation?

From the question the rate of the decay of the mass is proportional to the mass remaining at time #t#. And because the mass is decreasing with time the constant of proportionality will be negative and thus the differential equation can be expressed as ,

.#[a]#........#[dm]/dt=-km#, where #m#=mass and #k# is the constant of proportionality. we now need to integrate this expression,
inverting both ides of .......#[a]#,

#[dt]/[dm#=#-1/[km]#, integrating w.r.t #m#, #t=-1/klnm+C#......#[2]#.

Now let #m=m0# [ the initial mass when #t=0#]

So from .....#[2]# #0=-1/klnm0+C#, therefore #C=1/klnm0#

#t=-1/klnm+1/klnm0#, #t=1/k[lnm0-lnm]#

Therefore, #kt=ln[[mo]/[m]]# and so, #e^[kt]=[mo]/m# which will give #m=m0e^[-kt]#....and this is the answer to part #b#.

For part ...#[c]# we are given that #m0# [ intial mass =#125.3] and #m=98.1[ mass remaining when #t=3#]

And so from answer part #[b]#, #98.1=[125.3]e^[-3k]#

#98.1/125.3=e^[-3k]#, therefore -#1/3ln[98.1/125.3]#=k [ theory of logs]

When evaluated the above expression yields a value of #k#=#-0.081575#[approx answer to part #c#]

For part #[d]#, we now know that #m# =#m0e^-[0.081575t]# and we are given that #t=6# and we know the initial amount is #125.3# grammes.

Therefore #m=125.3e^-[6[0.081575]]# and when evaluated gives #m# the amount remaining =#76.8043# grammes[ approx]

Hope this was helpful.