Calculating the age by radioactive dating

The following was an example done in class, so I know the answer and (maybe?) the formula, but I still can't derive it for myself, and I would appreciate some help.



> 
> Given a sample of biotite with $\ce{K2O} = 9~\%,~~ \ce{^{40}Ar} = 2.242~\mathrm{\mu g/g}$, find the age.
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The constants given were $x\_e= 0.581 \cdot 10^{-10}\ \mathrm{y^{-1}}$, $x=5.543 \cdot 10^{-10}\ \mathrm{y^{-1}}$ where $x$ is the decay constant, isotopic abundance of $\ce{^{40}K} = 0.01167~\%$, and relative atomic masses $M\_\mathrm{r}(\ce{K}) = 39.098, M\_\mathrm{r}(\ce{O}) = 15.999, M\_\mathrm{r} (\ce{^{40}Ar}) =39.96$.


The answer shown was $2308 \cdot 10^6~\mathrm{y}$.


The formula I'm used to for this type of problem is $t=\frac{1}{x} \ln(\frac{\ce{^{40}Ar}^\*}{\ce{^{40}K}} (x/x\_e +1))$ but I'm not sure how to apply that given just a percentage of $\ce{K}$ instead of a mass. I thought of just finding the weight percentage of $\ce{K}$ from the $9~\%$ (its $80.46~\%$ of the $\ce{K\_2O}$, or $7.2~\%$ of the total), but that didn't work. I'm not sure if I need to consider atmospheric $\ce{Ar}$ either.


I tried $^{40}K = .09 \* .83022 \* .0001167 = 8.719 \*10^{-6}$ where .833022 is the weight percentage and it didn't give the correct answer.


I also tried $^{40}K = .09 \* .6666667 \* .0001167 = 7.002 \*10^{-6}$ using the molar percentage of K, but also didn't get the correct answer.


