Another Problem from Gaokao 2018

Here , $ab =\left(\dfrac{\log 3 - 1}{\log 2 -1}\right)\left(\dfrac{\log 3 -1}{\log 2}\right) = \dfrac{\,(\log 3 -1)^2 }{\log 2\,(\log 2 -1)} < 0$ Since $10 > 3 > 2 \implies \log 10 > \log 3 > \log 2\\ 1 > \log 3\implies 0 > \log 2 -1 \implies \log 2 -1 < 0\\ \therefore ab < 0$ Also $a+ b = \dfrac{\log 3 -1}{\log 2 -1} + \dfrac{\log 3 -1}{\log 2}= \,(\log 3 -1 ) \left ( \dfrac{\log 4 -1}{\log 2\,(\log 2 -1 ) }\right) <0 \\ \implies a+ b < 0$ Further $\dfrac{ab}{a+b} = \dfrac{\log 3 - 1}{\log 4 -1} > 1 \implies ab > a +b \\ 4 > 3 \implies \log 4 -1 > \log 3-1 \implies \dfrac{\log 3 -1}{\log 4 -1} > 1$ Therefore, $ab < 0 , a+b < 0 , ab > a+b \\ \implies ab < a+b < 0$