Question

Let $x$ and $y$ are solution of the system of equations $\{\begin{array}{c}{\log }_8\left(1\right)+{\log }_3(x+2)={\log }_3(3-2y)\\ 2^{y-x}-2^{3-y}=0\end{array}$ The value of $(y-x)$, is

• A. $2$
• B. $5$
• C. $11$
• D. $none$

Answer 1

The correct option is A $2$

$\begin{array}{c}{\log }_8^1+{\log }_3^{\left(x+2\right)}={\log }_3^{\left(3-2y\right)}\\ \Rightarrow {\log }_3^{\left(x+2\right)}={\log }_3^{\left(3-2y\right)}\\ \Rightarrow x+2=3-2y\\ \Rightarrow 2y=3-2-x\\ 2y=1-x\to \left(i\right)\\ and\Rightarrow 2^{\left(y-x\right)}=2^{\left(3-y\right)}\\ \Rightarrow y-x=3-y\\ 2y=3+x\to \left(ii\right)\\ fromequation1\&2\\ \Rightarrow 3+x=1-x\\ x=-1\\ andy=\frac{2}{2}=1\\ theny-x=1-(-1)=2\end{array}$