Express the equation $\log_2 x + \log_4(x-1) = 3$ as a cubic equation in $x$.

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

$\log_4(x-1) = \frac{\log_2(x-1)}{\log_2 4} = \frac{\log_2(x-1)}{2}$

2. Substitute back into the equation and eliminate the fraction

$\log_2 x + \frac{1}{2}\log_2(x-1) = 3$
$2\log_2 x + \log_2(x-1) = 6$

3. Combine the logarithmic terms using the power and product laws

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

4. Convert to exponential form and express as a cubic equation

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