differential equations A certain moon rock was found to contain equal numbers of potassium and argon atoms. Assume that all the argon is the result of radioactive decay of potassium (half life $1.28 \times 10^9$) and that one of every 9 potassium atom disintegrations yields an argon atom. What is the age of the rock measured from the time it contained only potassium.
please suggest
 A: After Rahul's comment above, I understand the question now. 
After $t$ years, exactly $\left(\frac{1}{2}\right)^{t/(1.28\times 10^9)}$ of the potassium will remain, and exactly $\frac{1}{9}$ of the decays will have produced an argon atom. Thus, we have that
$$\text{argon}=\frac{1}{9}\cdot\text{ decayed potassium}$$
and
$$\text{decayed potassium}=\text{original potassium} - \text{remaining potassium}$$
so
$$\text{decayed potassium}=\left(1-\left(\frac{1}{2}\right)^{t/(1.28\times 10^9)}\right)\cdot\text{ original potassium},$$
so
$$\text{argon}=\frac{1}{9}\cdot\left(1-\left(\frac{1}{2}\right)^{t/(1.28\times 10^9)}\right)\cdot\text{ original potassium}$$
Now we set this equal to the remaining amount of potassium, which is
$$\text{remaining potassium }=\left(\frac{1}{2}\right)^{t/(1.28\times 10^9)}\cdot\text{ original potassium}$$
This gives the equation
$$\frac{1}{9}\cdot\left(1-\left(\frac{1}{2}\right)^{t/(1.28\times 10^9)}\right)=\left(\frac{1}{2}\right)^{t/(1.28\times 10^9)}$$
in which we want to solve for $t$. This is just simple manipulations now.

We have
$$\frac{1}{9}\cdot\left(1-\left(\frac{1}{2}\right)^{t/(1.28\times 10^9)}\right)=\left(\frac{1}{2}\right)^{t/(1.28\times 10^9)}$$
which becomes
$$\left(\frac{1}{2}\right)^{t/(1.28\times 10^9)}=\frac{1}{10}$$
so $$\frac{t}{1.28\times 10^9}\times\ln\left(\frac{1}{2}\right)=\ln\left(\frac{1}{10}\right)$$
so
$$t=(1.28\times10^9)\frac{\ln\left(\frac{1}{10}\right)}{\ln\left(\frac{1}{2}\right)}=4.252\ldots\times 10^9$$
(see here)
