Show that the solution of the equation $3^{2x-1} = 2^{3-x}$ is $x = \frac{\lg 24}{\lg 18}$. Given that $\log_6 4y = \log_6 3x^2 - 1$, express $y$ in terms of $x$. Solve the equation $\log_2 x^2 = 8\log_x 2$.

1. Take the common logarithm (lg) of both sides of the equation

$3^{2x-1} = 2^{3-x}$
$\lg(3^{2x-1}) = \lg(2^{3-x})$

2. Apply the power law of logarithms to bring down the exponents

$(2x - 1)\lg 3 = (3 - x)\lg 2$

3. Expand both sides to isolate terms with x

$2x\lg 3 - \lg 3 = 3\lg 2 - x\lg 2$
$2x\lg 3 + x\lg 2 = 3\lg 2 + \lg 3$

4. Factorise x and apply logarithm laws to simplify the constants

$x(2\lg 3 + \lg 2) = \lg(2^3) + \lg 3$
$x(\lg(3^2) + \lg 2) = \lg 8 + \lg 3$
$x(\lg 9 + \lg 2) = \lg(8 \times 3)$
$x(\lg 18) = \lg 24$

5. Solve for x to obtain the final expression

$x = \frac{\lg 24}{\lg 18}$

1. Express the constant 1 as a logarithm with base 6

$\log_6 4y = \log_6 3x^2 - \log_6 6$

2. Apply the quotient law of logarithms to combine the terms on the right hand side

$\log_6 4y = \log_6 \left(\frac{3x^2}{6}\right)$
$\log_6 4y = \log_6 \left(\frac{x^2}{2}\right)$

3. Equate the arguments of the logarithms and solve for $y$

$4y = \frac{x^2}{2}$
$y = \frac{x^2}{8}$

1. Apply the power law and change of base formula to express the equation in terms of $\log_2 x$

$\log_2 x^2 = 8\log_x 2$
$2\log_2 x = 8 \left( \frac{1}{\log_2 x} \right)$

2. Form and solve a quadratic equation in $\log_2 x$

Let $u = \log_2 x$.
$2u = \frac{8}{u}$
$2u^2 = 8$
$u^2 = 4$
$u = 2 \text{ or } u = -2$

3. Solve for $x$ and check validity

$\log_2 x = 2 \implies x = 2^2 = 4$
$\log_2 x = -2 \implies x = 2^{-2} = \frac{1}{4}$
Both values are valid since $x > 0$ and $x \neq 1$.