Question

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

Answer 1

Answer: 1

$a={\log }_{12}18\&b={\log }_{24}54$
$$\begin{gathered} {\log }_{12}18\cdot {\log }_{24}54+5\left({\log }_{12}18-{\log }_{24}54\right) \\ <br/>{\log }_{8}27+5 \end{gathered}$$
$a=\frac{{\log }_{3}18}{{\log }_{3}12}=\frac{2+{\log }_{3}2}{1+2{\log }_{3}2}$
$b={\log }_{24}54$
$b=\frac{{\log }_{3}54}{{\log }_{3}24}=\frac{3+{\log }_{3}2}{1+3{\log }_{3}2}$
Let ${\log }_{3}2$ be $x$
$\therefore a=\frac{2+x}{1+2x},b=\frac{3+x}{1+3x}$
$\therefore ab+5(a-b)$
$=\frac{(2+x)(3+x)}{(1+2x)(1+3x)}+5\left[\frac{2+x}{1+2x}-\frac{3+x}{1+3x}\right]$
$=\frac{6+5x+x^2+5\left(2+6x+x+3x^2-3-6x-x-2x^2\right)}{(1+2x)(1+3x)}=\frac{6+5x+x^2+5\left(x^2-1\right)}{(1+2x)(1+3x)}$
$=\frac{x^2+5x+6+5x^2-5}{(1+2x)(1+3x)}=\frac{6x^2+5x+1}{6x^2+5x+1}=1$