Question

If ${\log }_{12}18=x,{\log }_{24}4=y$ , then the value of $xy-(5y+2x)+4$ will be

Answer 1

$x={\log }_{12}18=\frac{\log 18}{\log 12}$ ......... $[\because {\log }_{b}a=\frac{\log a}{\log b}]$
$=\frac{\log (2\cdot 3^2)}{\log (2^2\cdot 3)}=\frac{2\log 3+\log 2}{2\log 2+\log 3}$ ......... $[\because \log (ab)=\log a+\log b\text{and}\log a^m=m\log a]$
Similarly, $y={\log }_{24}4=\frac{2\log 2}{3\log 2+\log 3}$
Consider, $z=xy-(5y+2x)+4=(x-5)(y-2)-6$
$\Rightarrow z=(\frac{2\log 3+\log 2}{2\log 2+\log 3}-5)(\frac{2\log 2}{3\log 2+\log 3}-2)-6$
$\Rightarrow z=\left(\frac{2\log 3+\log 2-10\log 2-5\log 3}{2\log 2+\log 3}\right)\left(\frac{2\log 2-6\log 2-2\log 3}{3\log 2+\log 3}\right)-6$
$\Rightarrow z=\left(\frac{-3\log 3-9\log 2}{2\log 2+\log 3}\right)\left(\frac{-4\log 2-2\log 3}{3\log 2+\log 3}\right)-6=3\times 2-6=0$
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