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Let $$a = \log 2$$, $$b = \log 3$$ (any base).
$$p = \log_{18}24 = \dfrac{\log(2^3 \cdot 3)}{\log(2 \cdot 3^2)} = \dfrac{3a + b}{a + 2b}$$, so $$\dfrac{a}{b} = \dfrac{1 - 2p}{p - 3}$$.
$$\log_{96}108 = \dfrac{\log(2^2 \cdot 3^3)}{\log(2^5 \cdot 3)} = \dfrac{2a + 3b}{5a + b}$$. Substituting $$a/b$$ in terms of $$p$$ gives $$\dfrac{p+7}{9p-2}$$.
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