KVPY #18

$\begin{cases} a = \log_{245} 175 = \dfrac {\log_5 175}{\log_5 245} = \dfrac {\log_5 7 + 2\log_5 5}{2\log_5 7 + \log_5 5} = \dfrac {{\color{#3D99F6}\alpha}+2}{2{\color{#3D99F6}\alpha}+1} \\ b = \log_{1715} 875 = \dfrac {\log_5 875}{\log_5 1715} = \dfrac {\log_5 7 + 3\log_5 5}{3\log_5 7 + \log_5 5} = \dfrac {{\color{#3D99F6}\alpha}+3}{3{\color{#3D99F6}\alpha}+1} \end{cases} \quad \small \color{#3D99F6} \text{where }\alpha = \log_5 7$ .

$\begin{aligned} \implies \frac {1-ab}{a-b} & = \frac {1- \frac {\alpha+2}{2\alpha+1}\cdot \frac {\alpha+3}{3\alpha+1}}{\frac {\alpha+2}{2\alpha+1} - \frac {\alpha+3}{3\alpha+1}} \\ & = \frac {(2\alpha+1)(3\alpha+1) -(\alpha+2)(\alpha+3)}{(\alpha+2)(3\alpha+1) -(\alpha+3)(2\alpha+1)} \\ & = \frac {(6\alpha^2+5\alpha+1) -(\alpha^2+5\alpha+6)}{(3\alpha^2+7\alpha+2) -(2\alpha^2+7\alpha+3)} \\ & = \frac {5\alpha^2-5}{\alpha^2 - 1} = \boxed{5} \end{aligned}$