= Solution
Subtract the nepheline–corundum reaction from the albite–corundum reaction and divide by two. In the albite-to-nepheline direction the resulting net reaction is
$$
\mathrm{NaAlSi_3O_8}+4\mathrm{H_2O}
\rightleftharpoons\mathrm{NaAlSiO_4}+2\mathrm{H_4SiO_4}.
$$
Additivity of the <standard Gibbs energy of reaction> makes the logarithms of the <equilibrium constants> combine with the same coefficients:
$$
\boxed{\log_{10}K=\frac{-15.18-6.88}{2}=-11.03.}
$$
This normalization is per one <albite> formula unit. The doubled reaction has $\log_{10}K=-22.06$, and reversing the reaction changes its sign.
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