6 Solve the equation (i) $$ \quad 2 \log _{3}(x+4)-\log _{3}(x+2)=2 $$, (ii) $$ \quad \log _{5} y-3 \log _{y} 5=2 $$.

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Solve the equation $$ 2\log_{3}(x+4)-\log_{3}(x+2)=2 $$.
Solve the equation by following steps:
- step0: Solve for $$x$$:
  $$2\log_{3}{\left(x+4ight)}-\log_{3}{\left(x+2ight)}=2$$
- step1: Find the domain:
  $$2\log_{3}{\left(x+4ight)}-\log_{3}{\left(x+2ight)}=2,x\gt -2$$
- step2: Add the terms:
  $$\log_{3}{\left(\frac{\left(x+4ight)^{2}}{x+2}ight)}=2$$
- step3: Convert the logarithm into exponential form:
  $$\frac{\left(x+4ight)^{2}}{x+2}=3^{2}$$
- step4: Evaluate the power:
  $$\frac{\left(x+4ight)^{2}}{x+2}=9$$
- step5: Cross multiply:
  $$\left(x+4ight)^{2}=\left(x+2ight)	imes 9$$
- step6: Simplify the equation:
  $$\left(x+4ight)^{2}=9\left(x+2ight)$$
- step7: Expand the expression:
  $$x^{2}+8x+16=9\left(x+2ight)$$
- step8: Expand the expression:
  $$x^{2}+8x+16=9x+18$$
- step9: Move the expression to the left side:
  $$x^{2}+8x+16-\left(9x+18ight)=0$$
- step10: Calculate:
  $$x^{2}-x-2=0$$
- step11: Factor the expression:
  $$\left(x-2ight)\left(x+1ight)=0$$
- step12: Separate into possible cases:
  $$\begin{align}&x-2=0\&x+1=0\end{align}$$
- step13: Solve the equation:
  $$\begin{align}&x=2\&x=-1\end{align}$$
- step14: Check if the solution is in the defined range:
  $$\begin{align}&x=2\&x=-1\end{align},x\gt -2$$
- step15: Find the intersection:
  $$\begin{align}&x=2\&x=-1\end{align}$$
- step16: Rewrite:
  $$x_{1}=-1,x_{2}=2$$
Solve the equation $$ \log_{5}y-3\log_{y}5=2 $$.
Solve the equation by following steps:
- step0: Solve for $$y$$:
  $$\log_{5}{\left(yight)}-3\log_{y}{\left(5ight)}=2$$
- step1: Find the domain:
  $$\log_{5}{\left(yight)}-3\log_{y}{\left(5ight)}=2,y \in \left(0,1ight)\cup \left(1,+\inftyight)$$
- step2: Move the expression to the left side:
  $$\log_{5}{\left(yight)}-3\log_{y}{\left(5ight)}-2=0$$
- step3: Use the logarithm base change rule:
  $$\log_{5}{\left(yight)}-3	imes \frac{1}{\log_{5}{\left(yight)}}-2=0$$
- step4: Rewrite the expression:
  $$\log_{5}{\left(yight)}-\frac{3}{\log_{5}{\left(yight)}}-2=0$$
- step5: Solve using substitution:
  $$t-\frac{3}{t}-2=0$$
- step6: Multiply both sides of the equation by LCD:
  $$\left(t-\frac{3}{t}-2ight)t=0	imes t$$
- step7: Simplify the equation:
  $$t^{2}-3-2t=0$$
- step8: Factor the expression:
  $$\left(t-3ight)\left(t+1ight)=0$$
- step9: Separate into possible cases:
  $$\begin{align}&t-3=0\&t+1=0\end{align}$$
- step10: Solve the equation:
  $$\begin{align}&t=3\&t=-1\end{align}$$
- step11: Substitute back:
  $$\begin{align}&\log_{5}{\left(yight)}=3\&\log_{5}{\left(yight)}=-1\end{align}$$
- step12: Calculate:
  $$\begin{align}&y=125\&\log_{5}{\left(yight)}=-1\end{align}$$
- step13: Calculate:
  $$\begin{align}&y=125\&y=\frac{1}{5}\end{align}$$
- step14: Check if the solution is in the defined range:
  $$\begin{align}&y=125\&y=\frac{1}{5}\end{align},y \in \left(0,1ight)\cup \left(1,+\inftyight)$$
- step15: Find the intersection:
  $$\begin{align}&y=125\&y=\frac{1}{5}\end{align}$$
- step16: Rewrite:
  $$y_{1}=\frac{1}{5},y_{2}=125$$
The solutions to the equations are:
(i) $$ x = -1 $$ or $$ x = 2 $$
(ii) $$ y = \frac{1}{5} $$ or $$ y = 125 $$
The solutions are: (i) $$ x = -1 $$ or $$ x = 2 $$ (ii) $$ y = \frac{1}{5} $$ or $$ y = 125 $$

6 Solve the equation
(i) ,
(ii) .

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