Question

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

Answer 1

$a=\log _{12} 18 \& b=\log _{24} 54$
$\log _{12} 18 \cdot \log _{24} 54+5\left(\log _{12} 18-\log _{24} 54\right) \\
\log _8 27+5 $
$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+2 x}, b=\frac{3+x}{1+3 x} $
$\therefore a b+5(a-b)$
$=\frac{(2+x)(3+x)}{(1+2 x)(1+3 x)}+5\left[\frac{2+x}{1+2 x}-\frac{3+x}{1+3 x}\right] $
$=\frac{6+5 x+x^2+5\left(2+6 x+x+3 x^2-3-6 x-x-2 x^2\right)}{(1+2 x)(1+3 x)}=\frac{6+5 x+x^2+5\left(x^2-1\right)}{(1+2 x)(1+3 x)}$
$=\frac{x^2+5 x+6+5 x^2-5}{(1+2 x)(1+3 x)}=\frac{6 x^2+5 x+1}{6 x^2+5 x+1}=1$