If $\log_{18} 24 = p$, then $\log_{96} 108$ equals

Switch everything to base $2$ and base $3$. Let $L_2 = \log 2$ and $L_3 = \log 3$.

$\log 24 = \log(2^3 \cdot 3) = 3L_2 + L_3$

$\log 18 = \log(2 \cdot 3^2) = L_2 + 2L_3$

$p = \dfrac{3L_2 + L_3}{L_2 + 2L_3}$

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Similarly:

$\log 108 = \log(2^2 \cdot 3^3) = 2L_2 + 3L_3$

$\log 96 = \log(2^5 \cdot 3) = 5L_2 + L_3$

$\log_{96} 108 = \dfrac{2L_2 + 3L_3}{5L_2 + L_3}$

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Let $t = \dfrac{L_2}{L_3}$. Divide top and bottom of each fraction by $L_3$:

$p = \dfrac{3t + 1}{t + 2}$, which gives $t = \dfrac{2p - 1}{3 - p}$

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Substitute into the target expression:

$\log_{96} 108 = \dfrac{2t + 3}{5t + 1}$

Numerator $= \dfrac{2(2p-1)}{3-p} + 3 = \dfrac{4p - 2 + 9 - 3p}{3-p} = \dfrac{p + 7}{3-p}$

Denominator $= \dfrac{5(2p-1)}{3-p} + 1 = \dfrac{10p - 5 + 3 - p}{3-p} = \dfrac{9p - 2}{3-p}$

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The $(3-p)$ cancels:

$\log_{96} 108 = \dfrac{p + 7}{9p - 2}$

Answer $= \dfrac{p+7}{9p-2}$