If $\mathrm{a}={\log }_{245}175\text{ and }\mathrm{b}={\log }_{1715}875$ then the value of $\frac{1-\mathrm{ab}}{\mathrm{a}-\mathrm{b}}$ is

$\begin{array}{l}\mathrm{a}=\frac{{\log }_{5}175}{{\log }_{5}245}=\frac{2+{\log }_{5}7}{1+2{\log }_{5}7}\\ \text{ or }\mathrm{a}+2{\mathrm{alog}}_{5}7=2+{\log }_{5}7\\ \text{ or }{\log }_{5}7=\frac{\mathrm{a}-2}{1-2\mathrm{a}}----\left(\mathrm{i}\right)\end{array}$

$\text{ Now }\mathrm{b}=\frac{{\log }_{5}875}{{\log }_{5}1715}=\frac{3+{\log }_{5}7}{1+3{\log }_{5}7}$

$\begin{array}{r}\mathrm{or} \mathrm{b}+3{\mathrm{blog}}_{5}7=3+{\log }_{5}7\\ \mathrm{or} {\log }_{5}7=\frac{\mathrm{b}-3}{1-3\mathrm{b}}----\left(\mathrm{ii}\right)\end{array}$

From Eqs. (i) and (ii), we get 

$\begin{array}{r}\frac{\mathrm{a}-2}{1-2\mathrm{a}}=\frac{\mathrm{b}-3}{1-3\mathrm{b}}\\ \Rightarrow \frac{1-\mathrm{ab}}{\mathrm{a}-\mathrm{b}}=5\end{array}$