将方程 $\log_2 x + \log_4 (x-1) = 3$ 表示为关于 $x$ 的三次方程。

1. Change the base of the second logarithmic term to base 2.

$\log_2 x + \frac{\log_2 (x-1)}{\log_2 4} = 3$
$\log_2 x + \frac{\log_2 (x-1)}{2} = 3$

2. Multiply by 2 and use the power law of logarithms.

$2\log_2 x + \log_2 (x-1) = 6$
$\log_2 x^2 + \log_2 (x-1) = 6$

3. Combine the logarithmic terms using the product law.

$\log_2 [x^2(x-1)] = 6$

4. Convert to exponential form and expand to obtain a cubic equation.

$x^2(x-1) = 2^6$
$x^3 - x^2 = 64$
$x^3 - x^2 - 64 = 0$