Exponential decay involving logarithm

Question

In 2011 reactor $X$ released $4.2$ times the amount of cesium-137 as was leaked during reactor $Y$ disaster in 1986?

Using; A = Pert Half-life = $30.2$ years.

a) What year will cesium-137 level be half as much as immediately after the 2011 $X$ disaster?

b) In what year will cesium-137 level be a quarter as much as immediately after the 2011 $X$ disaster?

c) How long after the disaster until cesium-137 levels are the same as immediately after the 1986 $Y$ disaster?

a) Answer: 2041 one half-life
b) Answer: 2071 two half-life
c) Answer: 0 immediate would have same levels therefore it's zero.

Are my answers correct if not what am I doing wrong?

Answer 1

a and b are fine. For c you need to cut it by a factor 4.2.

Answer 2

Exponential decay follows the function $$N(t)=N_0 e^{kt}$$where $k$ is related to the half-life $t_{1/2}$ as $$N(t_{1/2})=\frac{N_0}{2}=N_0 e^{-kt_{1/2}} \leftrightarrow k=\frac{\ln(2)}{t_{1/2}}$$

(notice it is found as the time where the number of radioactive particles left are half of the original amount, namely $\frac{N_0}{2}$).

$a$ and $b$ look fine.

Now, for question $c)$, you want to find the time $t_c$ that the 2011 disaster (which was $4.2$ times as large as the 1986 one) has decayed to be at the same level as the 1986 disaster. This is the same as writing $$N(t_c)=\frac{N_0}{4.2}=N_0 e^{-kt_c}$$ which can be solved for $t_c$ using your knowledge of how $k$ is related to the half-life (which you know).