I have tried for days to figure out how to work this problem. Please help show me how to find the answer.

A particular radioactive isotope decays from 210 mg to 182.7 mg in 15 days.
Find the half life of the isotope.
I can not find any problem that is similar. everything i find is looking for a different part of the problem.

3 answers

Note on log:
At college and university level, log to the base 10 is rarely used, so often log stands for natural log, sometimes also written as ln.
Log10 is used for log to the base 10.
If you are not sure, check
ln(0.5)=log(0.5)=-0.6931472
log10(0.5)=-0.3010300

To start, we need to know the general formula for an exponential decay problem in terms of the material's decay constant, λ.

N(t)=N0*e^-λt
where
N0 is the quantity at time t=0
N(t) is the quantity remaining at time t.
e = Napier constant = 2.7182818284...
λ=decay constant.

For the given problem, we are given a set of decay parameters, from which we can calculate the value of λ, after which the quantity remaining at any time t can be determined.

N0=210, N(15)=182.7
so
182.7=210 e^{-λ (15)}
divide by 210 and take (natural) log on both sides:
log(182.7/210)=-15λ
Solving for λ gives
λ=-(1/15)log(182.7/210)
=0.009284 approximately

So the decay formula gives us
N(t)/N0 = e^-0.009284t

Half life is the time it takes for the decay to exactly half it's original quantity, so N(t)/N0=0.5, or
solve for t=half-life in:
(1/2) = e^-0.009284t
Again, take (natural) log on both sides,
-0.009284t = log(1/2)
or
half-life = t = -log(1/2) / 0.009284
=74.66 days

Thank you so much

You're welcome!