If  ${\log }_{12}18=\alpha$ and  ${\log }_{24}54=\beta$, then the value  $\alpha \beta +5(\alpha -\beta )=$.

$\alpha ={\log }_{12}18$

$=\frac{\log 18}{\log 12}$

$=\frac{\log (3^2\cdot 2)}{\log (3\cdot 2^2)}$

$=\frac{2\log 3+\log 2}{\log 3+2\log 2}$

$\beta ={\log }_{24}54$

$=\frac{\log 54}{\log 24}$

$=\frac{\log (3^3\cdot 2)}{\log (3\cdot 2^3)}$

$=\frac{3\log 3+\log 2}{\log 3+3\log 2}$

Let  $\log 2=x$ and  $\log 3=y$

$\therefore \alpha =\frac{x+2y}{2x+y},\beta =\frac{x+3y}{3x+y}$

Now,  $\alpha \beta +5(\alpha -\beta )=\left(\frac{x+2y}{2x+y}\right)\left(\frac{x+3y}{3x+y}\right)+5[\frac{x+2y}{2x+y}-\frac{x+3y}{3x+y}]$

$=\frac{(x^2+5xy+6y^2)+5(3x^2+xy+6xy+2y^2-2x^2-xy-6xy-3y^2)}{(6x^2+5xy+y^2)}$

$=\frac{(x^2+5xy+6y^2)+5(x^2-y^2)}{(6x^2+5xy+y^2)}=\frac{x^2+5xy+6y^2+5x^2-5y^2}{6x^2+5xy+y^2}$

$=\frac{6x^2+5xy+y^2}{6x^2+5xy+y^2}=1$