Solve the equation

1. Apply the power law of logarithms

$\log_3(x + 4)^2 - \log_3(x + 2) = 2$

2. Apply the quotient law of logarithms

$\log_3 \frac{(x + 4)^2}{x + 2} = 2$

3. Convert the logarithmic equation to exponential form

$\frac{(x + 4)^2}{x + 2} = 3^2$
$\frac{(x + 4)^2}{x + 2} = 9$

4. Form and solve the resulting quadratic equation

$(x + 4)^2 = 9(x + 2)$
$x^2 + 8x + 16 = 9x + 18$
$x^2 - x - 2 = 0$
$(x - 2)(x + 1) = 0$
$x = 2 \text{ or } x = -1$

5. Check the validity of the solutions

For $x = 2$, $x + 4 = 6 > 0$ and $x + 2 = 4 > 0$.
For $x = -1$, $x + 4 = 3 > 0$ and $x + 2 = 1 > 0$.
Both solutions are valid.

1. Apply the change of base formula to express all logarithms in the same base.

$\log_5 y - 3\left(\frac{1}{\log_5 y}\right) = 2$

2. Use substitution to form a quadratic equation.

Let $u = \log_5 y$.
$u - \frac{3}{u} = 2$
$u^2 - 2u - 3 = 0$

3. Solve the quadratic equation for $u$.

$(u - 3)(u + 1) = 0$
$u = 3$ or $u = -1$

4. Solve for $y$ and verify the solutions.

When $u = 3$:
$\log_5 y = 3 \implies y = 5^3 = 125$

When $u = -1$:
$\log_5 y = -1 \implies y = 5^{-1} = \frac{1}{5}$

Both $y = 125$ and $y = \frac{1}{5}$ are valid since $y > 0$ and $y \neq 1$.