In his famous paper Age of meteorites and the earth (1956), Patterson uses the following expressions:
Which I assume are the same as:
$$ \left(\frac{Pb^{206}}{Pb^{204}}\right)_P = \left(\frac{Pb^{206}}{Pb^{204}}\right)_I + \left({e^{\lambda_{238}T}-1}\right)\cdot\left(\frac{U^{238}}{Pb^{204}}\right)_P \tag{1}$$
$$ \left(\frac{Pb^{207}}{Pb^{204}}\right)_P = \left(\frac{Pb^{207}}{Pb^{204}}\right)_I + \frac{{e^{\lambda_{235}T}-1}}{\left(\frac{U^{238}}{U^{235}}\right)_P}\cdot\left(\frac{U^{238}}{Pb^{204}}\right)_P \tag{2}$$
He describes how the Canyon Diablo meteorite had such a small $U^{238}/Pb^{204}$ ratio ($0.025$) that "no observable change in the isotopic composition of lead could have resulted from radioactive decay after the meteorite was formed". Then he concludes that the meteorite's lead is representative of the isotopic composition of primordial lead at the time meteorites were formed.
My first issue with this is that we already have all the values necessary to determine the parameters in his expressions without assuming that the primordial lead isotopic composition was similar to that of the Canyon Diablo meteorite:
$$\begin{array}& \left(\frac{Pb^{206}}{Pb^{204}}\right)_{P,\text{ Canyon Diablo}} &= 9.46\\ \left(\frac{Pb^{207}}{Pb^{204}}\right)_{P,\text{ Canyon Diablo}} &= 10.34\\ \left(\frac{U^{238}}{Pb^{204}}\right)_{P,\text{ Canyon Diablo}} &= 0.025 \end{array}$$ $$\left(\frac{U^{238}}{U^{235}}\right)_P = 137.8$$ $$\begin{array}& \lambda_{238} &= 1.537\times 10^{-10}\text{ yr}^{-1}\\ \lambda_{235} &= 9.72\times 10^{-10}\text{ yr}^{-1}\\ \end{array}$$ $$T = 4.55\times 10^9\text{ yr}$$
Plugging those values into equations $(1)$ and $(2)$ and solving for $\left(Pb^{206}/Pb^{204}\right)_I$ and $\left(Pb^{207}/Pb^{204}\right)_I$, we arrive at:
$$ \left(\frac{Pb^{206}}{Pb^{204}}\right)_P = 9.43 + 1.012\left(\frac{U^{238}}{Pb^{204}}\right)_P \tag{3}$$
$$ \left(\frac{Pb^{207}}{Pb^{204}}\right)_P = 10.33 + 0.597\left(\frac{U^{238}}{Pb^{204}}\right)_P \tag{4}$$
As you can see, the parameters are very similar to those in Patterson's expressions, but not exactly the same. The second problem is that directly substituting the approximate values below also doesn't work:
$$\begin{array}& \left(\frac{Pb^{206}}{Pb^{204}}\right)_I \approx \left(\frac{Pb^{206}}{Pb^{204}}\right)_{P,\text{ Canyon Diablo}} &= 9.46\\ \left(\frac{Pb^{207}}{Pb^{204}}\right)_I \approx \left(\frac{Pb^{207}}{Pb^{204}}\right)_{P,\text{ Canyon Diablo}} &= 10.34\\ \end{array}$$
So how exactly did he arrive at the parameters $9.50$, $1.014$, $10.36$, and $0.601$?
