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 ${\log }_{2013}\left(y_1{y}_2\right)=0$, $\Rightarrow y_1{y}_2=2013^0=1\Rightarrow {\log }_4\left(x_1{x}_2{y}_1{y}_2\right)=24$