Question

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

Answer 1

Given $\frac{\log \alpha }{(3x-5y-z)}=\frac{\log \beta }{(x+8z)}=\frac{\log \gamma }{(y-3z-x)}$
$\frac{\lambda \log \alpha }{\lambda (3x-5y-z)}=\frac{2\lambda \log \beta }{2\lambda (x+8z)}=\frac{5\lambda \log \gamma }{5\lambda (y-3z-x)}=k$
On comparing we get,
$\log {\alpha }^{\lambda }=k\lambda (3x-5y-z)$...(1)
$\log {\beta }^{2\lambda }=2k\lambda (x+8z)$ ....(2)
$\log {\gamma }^{5\lambda }=5k\lambda (y-3z-x)$ ....(3)
Adding (1),(2), (3), we get
$\log {\alpha }^{\lambda }+\log {\beta }^{2\lambda }+\log {\gamma }^{5\lambda }=k\left[\lambda \right(3x-5y-z)+2\lambda (x+8z)+5\lambda (y-3z-x)$
$\Rightarrow \log ({\alpha }^{\lambda }\cdot {\beta }^{2\lambda }\cdot {\gamma }^{5\lambda })=0$
$\Rightarrow (\alpha {\beta }^2{\gamma }^5{)}^{\lambda }=1$
Now, given ${\log }_{\alpha }a=2$
$\Rightarrow a={\alpha }^2$
Also given, ${\log }_{2\beta }2b=4$
$\Rightarrow 2b=2^4{\beta }^4$
Also given ${\log }_{4{\gamma }^2}16c=5$
$\Rightarrow 16c=(4{\gamma }^2{)}^5$
Now, consider $\left(\frac{a}{8}\right)\left(\frac{b}{4}\right)\left(\frac{c}{2}\right)$
$=2^3(\alpha {\beta }^2{\gamma }^5{)}^2=8$