Question

Let $\alpha ,\beta ,\gamma$ be positive integers and $\frac{\log \alpha }{(3x-5y-z)}=\frac{\log \beta }{(x+8z)}=\frac{\log \gamma }{(y-3z-x)}$ (wherever defined). If ${\log }_{\alpha }a=2,$ ${\log }_{2\beta }2b=4,$ ${\log }_{4{\gamma }^2}16c=5(a,b,c>0),$ then value of $\left(\frac{a}{8}\right)\left(\frac{b}{4}\right)\left(\frac{c}{2}\right)$is

Answer 1

$\frac{\log \alpha }{(3x-5y-z)}=\frac{\log \beta }{(x+8z)}=\frac{\log \gamma }{(y-3z-x)}=k\Rightarrow \frac{\log \alpha }{k}=3x-5y-z\cdots \left(1\right)\Rightarrow \frac{\log \beta }{k}=x+8z\cdots \left(2\right)\Rightarrow \frac{\log \gamma }{k}=-x+y-3z\cdots \left(3\right)$

Now, ${\log }_{\alpha }a=2,{\log }_{2\beta }2b=4,{\log }_{4{\gamma }^2}16c=5$
$\Rightarrow a={\alpha }^2,2b=(2\beta {)}^4,16c=(4y^2{)}^5\Rightarrow a={\alpha }^2,b=2^3{\beta }^4,c=2^6{y}^{10}$
Therefore, $\left(\frac{a}{8}\right)\left(\frac{b}{4}\right)\left(\frac{c}{2}\right)=\frac{abc}{2^6}=2^3\times (\alpha {\beta }^2{\gamma }^5{)}^2$
From equation $\left(1\right),\left(2\right)$ and $\left(3\right)$, we get
$\frac{\log \alpha }{k}+2\frac{\log \beta }{k}+5\frac{\log \gamma }{k}=3x-5y-z+2(x+8z)+5(-x+y-3z)\Rightarrow \frac{\log \left(\alpha {\beta }^2{\gamma }^5\right)}{k}=0\Rightarrow \alpha {\beta }^2{\gamma }^5=1$

Hence, $\left(\frac{a}{8}\right)\left(\frac{b}{4}\right)\left(\frac{c}{2}\right)=8$