= Solution
Let $K(t)$ be the surviving <potassium-40> population and $A_r(t)$ the accumulated radiogenic <argon-40>. The total <nuclear decay constant> $\lambda$ sets $K(t)=K_0e^{-\lambda t}$. Only fraction $b=0.11$ of lost potassium atoms produces argon, so under the stated retention and zero-initial-argon assumptions,
$$
A_r(t)=b[K_0-K(t)].
$$
Dividing by the present potassium abundance eliminates the unknown initial population:
$$
\frac{A_r}{K}=b(e^{\lambda t}-1),\qquad
\boxed{t=\frac{T_{1/2}}{\ln2}\ln\left(1+\frac{A_r/K}{b}\right).}
$$
With $A_r/K=0.3$,
$$
t=\frac{1.3\times10^9\,\mathrm{yr}}{\ln2}
\ln\left(1+\frac{0.3}{0.11}\right)
=2.4676\times10^9\,\mathrm{yr}.
$$
Thus the inferred <potassium-argon dating> age is \b[about $2.5$ billion years], consistent with the proposed $2.5\times10^9$-year age to the accuracy of the supplied inputs. Omitting the branching fraction would seriously underestimate the age. At exactly $2.5$ billion years, the model predicts $A_r/K\simeq0.307$, consistent with the rounded measured $0.3$.
This dates closure of the mineral to argon loss and potassium exchange under the assumptions, not necessarily its original crystallization or the first formation of the Earth's crust. The depth alone is not an age measurement. No uncertainty or argon-retention history is supplied, so the agreement is a consistency check rather than proof of a particular geological event.
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