Question

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

Answer 1

$a={\log }_{12}18=\frac{\log 18}{\log 12}=\frac{\log (3^2\cdot 2)}{\log (2^2\cdot 3)}=\frac{2\log 3+\log 2}{2\log 2+\log 3}$
$[\because {\log }_{b}a=\frac{\log a}{\log b},\log (ab)=\log a+\log b,\log a^m=m\log a]$
Similarly $b={\log }_{24}54=\frac{\log 2+3\log 3}{3\log 2+\log 3}$
Now, $c=ab+5(a-b)=(a-5)(b+5)+25$
$\Rightarrow c=(\frac{2\log 3+\log 2}{2\log 2+\log 3}-5)(\frac{\log 2+3\log 3}{3\log 2+\log 3}+5)+25$
$\Rightarrow c=(\frac{2\log 3+\log 2}{2\log 2+\log 3}-5)(\frac{\log 2+3\log 3}{3\log 2+\log 3}+5)+25$
$\Rightarrow c=\left(\frac{-3\log 3-9\log 2}{2\log 2+\log 3}\right)\left(\frac{16\log 2+8\log 3}{3\log 2+\log 3}\right)+25=-24+25=1$
Ans: 1