If $\mathrm{a}={\log }_{12}18,\mathrm{b}={\log }_{24}54$ then the value of $\mathrm{ab}+5(\mathrm{a}-\mathrm{b})$ is

 we have $\mathrm{a}={\log }_{12}18=\frac{{\log }_{2}18}{{\log }_{2}12}=\frac{1+2{\log }_{2}3}{2+{\log }_{2}3}$

and $\mathrm{b}={\log }_{24}54=\frac{{\log }_{2}54}{{\log }_{2}24}=\frac{1+3{\log }_{2}3}{3+{\log }_{2}3}$

Putting $\mathrm{x}={\log }_{2}3\text{, we have }$

$\begin{array}{c}\mathrm{ab}+5(\mathrm{a}-\mathrm{b})=\frac{1+2\mathrm{x}}{2+\mathrm{x}}\cdot \frac{1+3\mathrm{x}}{3+\mathrm{x}}+5\left(\frac{1+2\mathrm{x}}{2+\mathrm{x}}-\frac{1+3\mathrm{x}}{3+\mathrm{x}}\right)\\ =\frac{6{\mathrm{x}}^2+5\mathrm{x}+1+5\left(-{\mathrm{x}}^2+1\right)}{(\mathrm{x}+2)(\mathrm{x}+3)}\\ =\frac{{\mathrm{x}}^2+5\mathrm{x}+6}{(\mathrm{x}+2)(\mathrm{x}+3)}=1\end{array}$