Question

If the solutions to the system of equations given by $\log _{4096} x+\log _{2013} y=2$ and $\log _{ x } 4096-\log _{ y } 2013=1$ are $\left( x _1, y _1\right)$ and $\left( x _2, y _2\right)$, then the value of $\log _4\left( x _1 y _1 x _2 y _2\right)$ is

• A. 36
• B. 30
• C. 24
• D. 18

Answer 1

Answer: (C)

Let $u =\log _{4096} x$ and $v =\log _{2013} y$ so that the equations becomes $u + v =2$ and $\frac{1}{ u }-\frac{1}{ v }=1$.
This system of solutions $(u, v)=(2 \mu \sqrt{2}, \pm \sqrt{2})$,
so that $u _1+ u _2=\log _{4096} x _1+\log _{4096} x _2=\log _{4096}\left( x _1 x _2\right)=4$ and $v _1+ v _2=\log _{2013} y _1+\log _{2013} y _2=\log _{2013}\left( y _1 y _2\right)=0$.
By change of base, if $\log _{4096}\left( x _1 x _2\right)=4$, then $\log _4\left( x _1 x _2\right)=24$ and if $\log _{2013}\left( y _1 y _2\right)=0$, $\Rightarrow y _1 y _2=2013^0=1 \Rightarrow \log _4\left( x _1 x _2 y _1 y _2\right)=24$