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)