Given that $\frac{(\log_y x)^2}{\log_x y} + 64 = 0$, express $y$ in terms of $x$. Solve the equation $\log_2 x + 3 = 2\log_2(x - 1)$.

1. Apply the change of base formula to express the equation in terms of a single logarithmic base.

Using the change of base formula, $\log_x y = \frac{1}{\log_y x}$.
Substitute this into the given equation:
$\frac{(\log_y x)^2}{\frac{1}{\log_y x}} + 64 = 0$

2. Simplify the equation and solve for $\log_y x$.

$(\log_y x)^3 + 64 = 0$
$(\log_y x)^3 = -64$
$\log_y x = -4$

3. Convert the logarithmic equation to exponential form to express $y$ in terms of $x$.

$y^{-4} = x$
$y = x^{-\frac{1}{4}}$ (or $y = \frac{1}{\sqrt[4]{x}}$)

1. Express all terms as logarithms with base 2 and apply the power law

$\log_2 x + \log_2 2^3 = \log_2(x - 1)^2$
$\log_2 x + \log_2 8 = \log_2(x - 1)^2$

2. Apply the product law of logarithms and equate the arguments

$\log_2(8x) = \log_2(x - 1)^2$
$8x = (x - 1)^2$

3. Expand and form a quadratic equation

$8x = x^2 - 2x + 1$
$x^2 - 10x + 1 = 0$

4. Solve the quadratic equation and reject the invalid root

$x = \frac{-(-10) \pm \sqrt{(-10)^2 - 4(1)(1)}}{2(1)}$
$x = \frac{10 \pm \sqrt{96}}{2}$
$x = \frac{10 \pm 4\sqrt{6}}{2}$
$x = 5 \pm 2\sqrt{6}$

For $\log_2(x - 1)$ to be defined, $x > 1$.
Since $5 - 2\sqrt{6} \approx 0.101 < 1$, we reject $x = 5 - 2\sqrt{6}$.
Therefore, $x = 5 + 2\sqrt{6}$.