Question

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

Answer 1

We have $a={\log }_{12}18=\frac{{\log }_{2}18}{{\log }_{2}12}=\frac{{\log }_2(2\times 3^2)}{{\log }_2(2^2\times 3)}=\frac{1+2{\log }_{2}3}{2+{\log }_{2}3}$

And $b={\log }_{24}54=\frac{{\log }_{2}54}{{\log }_{2}24}=\frac{{\log }_2(2\times 3^3)}{{\log }_2(2^3\times 3)}=\frac{1+3{\log }_{2}3}{3+{\log }_{2}3}$

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

$ab+5(a-b)=\frac{1+2x}{2+x}.\frac{1+3x}{3+x}+5(\frac{1+2x}{2+x}-\frac{1+3x}{3+x})$

$=\frac{{6x}^2+5x+1+(-x^2+1)}{(x+2)(x+3)}$

$=\frac{x^2+5x+6}{(x+2)(x+3)}=1$