So i'm told that a radioactive substance has a half life of 10 years and is modelled by the following equation:
A=A_0 * e^(-kt)
where A_0 is the original activity and k is some constant. And t is time in years. And i am given that time is 5 years and new activity is 200 Bq. How should i find the original activity?
If the half life is $10$ years, then $$A=A_0 2^{-t/10}$$ Then it is given that $$200=A_0 2^{-5/10}=\frac{A_0}{\sqrt{2}}$$ So $$A_0 = 200\sqrt{2} \text{ Bq}$$
Assuming that the time $t$ in the formula is measured in years, the information about the half-life tells us that at time $t=10$ the value is half that at time $t=0$, i.e., $$A_0e^{-10k}=\frac12A_0.$$ You can solve this equation for $k$ by eliminating the common factor of $A_0$ and taking the logarithm of both sides: $$-10k=\ln\frac12=-\ln2 \\ k = \frac1{10}\ln2$$ so that the decay formula becomes $$A_0e^{-\left(\frac1{10}\ln2\right)t}=A_02^{-t/10}.$$ The second fact tells you that $A_02^{-5/10}=200$. I’m sure that you can solve this equation for the initial value $A_0$ yourself.
"Explain why" about what? Do you know what "half life" means? If something has a "half life" of 10 years, and the initial amount is $A_0$, then in 10 years, the amount will be half that: $\left(\frac{1}{2}\right)A_0$. In another 10 years, 20 years, it will be half of that- $\frac{1}{2}\left(\frac{1}{2}\right)A_0= \frac{1}{2^2}A_0$. In another 10 years, 30 years, 1/2 of that- $\frac{1}{2}\left(\frac{1}{2^2}\right)A_0= \frac{1}{2^3}A_0$, etc. That powers of 2 are $\frac{10}{10}= 1$, $\frac{20}{10}= 2$, $\frac{30}{10}= 3$. After "t" years, that power of 2 will be $\frac{t}{10}$ so we have $\frac{1}{2^{t/10}}= 2^{-t/10}$
