Given that $2^{x-2} \times 3^{x+2} = 6^{2x}$, show that $x = \frac{\lg 9 - \lg 4}{\lg 6}$. Solve $\log_9 y - 2 = \log_3 2y$.

1. Take logarithms to base 10 on both sides of the equation.

$\lg(2^{x-2} \times 3^{x+2}) = \lg(6^{2x})$

2. Apply the product and power laws of logarithms to expand the equation.

$\lg(2^{x-2}) + \lg(3^{x+2}) = 2x \lg 6$
$(x-2)\lg 2 + (x+2)\lg 3 = 2x \lg 6$

3. Expand the terms and group the terms involving $x$.

$x \lg 2 - 2 \lg 2 + x \lg 3 + 2 \lg 3 = 2x \lg 6$
$x(\lg 2 + \lg 3) - \lg(2^2) + \lg(3^2) = 2x \lg 6$
$x \lg(2 \times 3) - \lg 4 + \lg 9 = 2x \lg 6$
$x \lg 6 - \lg 4 + \lg 9 = 2x \lg 6$

4. Rearrange to make $x$ the subject.

$\lg 9 - \lg 4 = 2x \lg 6 - x \lg 6$
$\lg 9 - \lg 4 = x \lg 6$
$x = \frac{\lg 9 - \lg 4}{\lg 6}$

1. Change the base of $\log_9 y$ to base 3

$\log_9 y = \frac{\log_3 y}{\log_3 9} = \frac{\log_3 y}{2}$

2. Substitute into the equation and eliminate fractions

$\frac{1}{2}\log_3 y - 2 = \log_3 2y$
$\log_3 y - 4 = 2\log_3 2y$

3. Apply power and quotient laws of logarithms to combine terms

$\log_3 y - 4 = \log_3 (2y)^2$
$\log_3 y - 4 = \log_3 (4y^2)$
$-4 = \log_3 (4y^2) - \log_3 y$
$-4 = \log_3 \left(\frac{4y^2}{y}\right)$
$-4 = \log_3 (4y)$

4. Convert to exponential form and solve for $y$

$4y = 3^{-4}$
$4y = \frac{1}{81}$
$y = \frac{1}{324}$