Question

If ${\log }_3\left(x\right)+{\log }_3\left(y\right)=2+{\log }_{3}2$ and ${\log }_3(x+y)=2$, then

• A. $x=1,y=8$
• B. $x=8,y=1$
• C. $x=3,y=6$
• D. $x=9,y=3$

Answer 1

The correct option is C

$x=3,y=6$

Explanation for the correct answer:

Step 1: Simplify the equations using log properties:

Given: ${\log }_3\left(x\right)+{\log }_3\left(y\right)=2+{\log }_{3}2$ and ${\log }_3(x+y)=2$

Consider ${\log }_3\left(x\right)+{\log }_3\left(y\right)=2+{\log }_{3}2$

$$\begin{gathered} \Rightarrow {\log }_3\left(xy\right)=2{\log }_3\left(3\right)+{\log }_3\left(2\right)\left[\because log\left(ab\right)=log\left(a\right)+log\left(b\right);lo{g}_a\left(a\right)=1\right] \\ \Rightarrow {\log }_3\left(xy\right)={\log }_3{\left(3\right)}^2+{\log }_3\left(2\right)\left[\because alog\left(b\right)=log{\left(b\right)}^a\right] \\ \Rightarrow {\log }_3\left(xy\right)={\log }_3\left(9\right)+{\log }_3\left(2\right) \\ \Rightarrow {\log }_3\left(xy\right)={\log }_3\left(18\right) \\ \Rightarrow xy=18...\left(1\right) \end{gathered}$$

$$\begin{gathered} {\log }_3(x+y)=2 \\ \Rightarrow (x+y)=3^2\left[\because {\log }_a\left(b\right)=c\Rightarrow b=a^c\right] \\ \Rightarrow x+y=9...\left(2\right) \end{gathered}$$

Step 2: Solve for the required value:

$x=9-y$

Put the value of $x$ in $\left(1\right)$

$$\begin{gathered} \Rightarrow (9-y)y=18 \\ \Rightarrow 9y-y^2=18 \\ \Rightarrow y^2-9y+18=0 \\ \Rightarrow y^2-6y-3y+18=0 \\ \Rightarrow y(y-6)-3(y-6)=0 \\ \Rightarrow (y-6)(y-3)=0 \\ \Rightarrow y=6,y=3 \end{gathered}$$

Consider,

$$\begin{gathered} y=6\Rightarrow x=9-y=9-6=3 \\ y=3\Rightarrow x=9-y=9-3=6 \end{gathered}$$

According to the given options the correct combination is $x=3,y=6$

Hence option(C) i.e. $x=3,y=6$ is correct.