= Solution
After complete <ionization>, each <hydrogen> nucleus supplies one <ion> and one <electron>, while each <helium-4> nucleus supplies one <ion> and two <electrons>. Neglecting the small <metal mass fraction> $Z$ and putting $Y\simeq1-X$, the <mean molecular weight> therefore satisfies
$$
\frac1\mu=2X+\frac34Y\simeq\frac{3+5X}{4},\qquad \boxed{\mu\simeq\frac4{3+5X}}.
$$
Use <stellar homology> to compare models with common dimensionless profiles. The <stellar mass conservation equation>, <stellar hydrostatic equation> and <ideal gas law> give
$$
\rho_c\propto\frac M{R^3},\qquad P_c\propto\frac{GM^2}{R^4},\qquad T_c\propto\frac{\mu M}{R}.
$$
The <radiative diffusion in a star> equation then gives $L_{\rm rad}\propto RT_c^4/(\kappa_c\rho_c)$. Inserting the specified <opacity> yields $L_{\rm rad}\propto RT_c^7/(Z\rho_c^2)\propto\mu^7M^5/Z$. Independently, integrating the <stellar energy-generation rate> over the <mass> gives $L_{\rm nuc}\propto X^2\rho_cT_c^5M\propto X^2\mu^5M^7/R^8$. Equating these in <stellar thermal equilibrium> gives the <radiative homology with fifth-power hydrogen burning and inverse-cubic opacity>:
$$
R^8\propto X^2Z\mu^{-2}M^2,\qquad R\propto X^{1/4}Z^{1/8}\mu^{-1/4}M^{1/4},\qquad T_c\propto\frac{\mu^{5/4}M^{3/4}}{X^{1/4}Z^{1/8}}.
$$
At fixed <solar mass> these reduce to \b[$L\propto\mu^7/Z$ and $T_c\propto\mu^{5/4}/(X^{1/4}Z^{1/8})$.] The <stellar homology> approximation, rather than merely the local power laws, is what permits these relations between whole stars.
Equal <luminosities> imply $\mu_2/\mu_1=(Z_2/Z_1)^{1/7}=2^{-1/7}$. Thus
$$
3+5X_2=2^{1/7}(3+5X_1),\qquad X_2=\frac{6.5\,2^{1/7}-3}{5}=0.8353\ldots,
$$
so \b[$X_2=0.8$ to one significant figure]. Retaining the unrounded <hydrogen mass fraction> when comparing <central temperatures> gives
$$
\frac{T_{c,2}}{T_{c,1}}=2^{-3/56}\left(\frac{X_1}{X_2}\right)^{1/4}=0.9219\ldots<1.
$$
\b[The first model has the higher central temperature.]
For homogeneous <hydrogen burning> at fixed $M,Z$, fuel conservation gives $ME_0\dot X=-L$. Since $d\mu/dX=-5\mu^2/4$ and $L=L_0(\mu/\mu_0)^7$, the <homogeneous fuel-depletion luminosity feedback> obeys
$$
\dot\mu=\frac{5L_0}{4ME_0\mu_0^7}\mu^9,\qquad \frac{d}{dt}\mu^{-8}=-\frac{10L_0}{ME_0\mu_0^7}.
$$
Integration with the initial <mean molecular weight> $\mu_0$ gives
$$
\boxed{L(t)=L_0\left(1-\frac{10\mu_0L_0t}{ME_0}\right)^{-7/8}}.
$$
This idealized <stellar evolution> law stops when $X=0$; it does not predict a physically attainable infinite <luminosity>. Indeed $\mu=4/3$ at fuel exhaustion, giving $t_H=ME_0[1-(3\mu_0/4)^8]/(10\mu_0L_0)$, strictly before the formal singularity.
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