Question

Given that, ${\log }_{a}x{\log }_a\left(xyz\right)=48$
${\log }_{a}y{\log }_a\left(xyz\right)=12,a>0,a\ne 1$
${\log }_{a}z{\log }_a\left(xyz\right)=84$
The solution(s) of the system of equations is/are

• A. only $(a^4,a,a^7)$
• B. only $(\frac{1}{a^7},\frac{1}{a},\frac{1}{a^4})$
• C. $(\frac{1}{a^4},\frac{1}{a},\frac{1}{a^7})$ or $(a^4,a,a^7)$
• D. $(a^7,a,a^4)$ or $(\frac{1}{a^7},\frac{1}{a},\frac{1}{a^4})$

Answer 1

The correct option is C $(\frac{1}{a^4},\frac{1}{a},\frac{1}{a^7})$ or $(a^4,a,a^7)$

Given ${\log }_{a}x{\log }_a\left(xyz\right)=48\to \left(1\right)$
${\log }_{a}y{\log }_a\left(xyz\right)=12,a>0,a\ne 1\to \left(2\right)$
${\log }_{a}z{\log }_a\left(xyz\right)=84\to \left(3\right)$
$\frac{\left(1\right)}{\left(2\right)}$and$\frac{\left(3\right)}{\left(2\right)}$ gives
$\frac{{\log }_{a}x}{{\log }_{a}y}=4$ & $\frac{{\log }_{a}z}{{\log }_{a}y}=7$
$\Rightarrow \frac{{\log }_{a}x}{4}=\frac{{\log }_{a}y}{1}=\frac{{\log }_{a}z}{7}=\lambda$
$\Rightarrow x=a^{4\lambda },y=a^{\lambda },z=a^{7\lambda }$
Now, ${\log }_a{a}^{4\lambda }{\log }_a\left(a^{4\lambda +\lambda +7\lambda }\right)=48$
$48{\lambda }^2=48\Rightarrow {\lambda }^2=1\Rightarrow \lambda =\pm 1$
$x=a^{\pm 4},y=a^{\pm 1},z=a^{\pm 7}$