$\text{Without using table, show that }\frac{\log {{8}^{\tfrac{2}{3}}}+\log {{729}^{\tfrac{1}{3}}}-\log {{125}^{\tfrac{2}{3}}}}{\log 6-\log 5}=2$
$\begin{align} & \frac{\log {{8}^{\tfrac{2}{3}}}+\log {{729}^{\tfrac{1}{3}}}-\log {{125}^{\tfrac{2}{3}}}}{\log 6-\log 5}=\frac{\log {{({{2}^{3}})}^{\tfrac{2}{3}}}-\log {{({{3}^{6}})}^{\tfrac{1}{3}}}-\log {{({{5}^{3}})}^{\tfrac{2}{3}}}}{\log 6-\log 5} \\ & \frac{\log {{8}^{\tfrac{2}{3}}}+\log {{729}^{\tfrac{1}{3}}}-\log {{125}^{\tfrac{2}{3}}}}{\log 6-\log 5}=\frac{\log {{2}^{2}}+\log {{3}^{2}}-\log {{5}^{2}}}{\log 6-\log 5} \\ & \frac{\log {{8}^{\tfrac{2}{3}}}+\log {{729}^{\tfrac{1}{3}}}-\log {{125}^{\tfrac{2}{3}}}}{\log 6-\log 5}=\frac{2\log 2+2\log 3-2\log 5}{\log 6-\log 5} \\ & \frac{\log {{8}^{\tfrac{2}{3}}}+\log {{729}^{\tfrac{1}{3}}}-\log {{125}^{\tfrac{2}{3}}}}{\log 6-\log 5}=\frac{2(\log 2+\log 3-\log 5)}{\log 6-\log 5} \\ & \frac{\log {{8}^{\tfrac{2}{3}}}+\log {{729}^{\tfrac{1}{3}}}-\log {{125}^{\tfrac{2}{3}}}}{\log 6-\log 5}=\frac{2(\log 6-\log 5)}{\log 6-\log 5} \\ & \frac{\log {{8}^{\tfrac{2}{3}}}+\log {{729}^{\tfrac{1}{3}}}-\log {{125}^{\tfrac{2}{3}}}}{\log 6-\log 5}=2 \\\end{align}$