differential equation12 After 30 days of radioactive decay,100 mg of a radioactive substance was observed to remain. After 120 days, only 30 mg of this substance was left. 
A. How much substance was originally present?
B. What is the half-life of this radioactive substance?
C. How long will it take before only 1% of the original amount remains?
I have no clue what so ever
 A: I will get you kicked off. As mentioned in the comments, we use an exponential for these types of problems, so we have:
$$A(t) = ce^{kt}$$
Now all you need to do is use the conditions specified in the problem to find the particular solution. 


*

*Using $t = 30$ and $A(30) = 100$, we get: $\displaystyle 100 = ce^{30k}$.

*Using $t = 120$ and $A(120) = 30$, we get: $\displaystyle 30 = ce^{120k}$.


The resulting system of equations is:
$$100 = ce^{30k}$$
$$30 = ce^{120k}$$
Using the first equation, we get:
$$\displaystyle c = \frac{100}{e^{30k}} = 100e^{-30k}$$
Substituting into the second equation, you have: $30 = 100e^{-30k} e^{120k} = 100e^{90k}$, or, $\frac{30}{100} = e^{90k}$, so
$$\displaystyle \ln\left(\frac{30}{100}\right) = 90k \rightarrow k = -0.013377$$
Notice here that $k$ is in fact negative. This is what you would expect given that we are talking about decay rather than growth.
Using the equation for $c$, we have: 
$$\displaystyle c = 100e^{-30 \times -0.013377} = 149.38$$
The amount $A$ as a function of time is given by:
$$\displaystyle A(t) = 149.38e^{-0.013377t}$$
To find the original amount of substance, use $t=0$, and we get:
$$A(0) = 149.38e^0 = 149.38~~ \text{mg}$$
How long will it take before only 1% of the "original amount remains"?
$$149.38 e^{-0.013377t} = .01 \times 149.38 \rightarrow t = 344.26 ~~ \text{days}.$$
