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

We have 

$\begin{array}{l}a={\log }_{12}18=\frac{{\log }_{2}18}{{\log }_{2}12}=\frac{1+2{\log }_{2}3}{2+{\log }_{2}3}\text{ and }\\ b={\log }_{24}54=\frac{{\log }_{2}54}{{\log }_{2}24}=\frac{1+3{\log }_{2}3}{3+{\log }_{2}3}\end{array}$

Putting $x={\log }_{2}3$ we have

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