Solve the logarithmic equation 4(x−1) = 2(x−3)

The given equation is $\log_4(x-1) = \log_2(x-3)$.
For the logarithms to be defined, their arguments must be positive:
$x-1 > 0 \implies x > 1$
$x-3 > 0 \implies x > 3$
Combining these conditions, the valid domain for $x$ is $x > 3$.

Use the change of base formula $\log_b a = \frac{\log_c a}{\log_c b}$ to convert $\log_4(x-1)$ to base 2:

Substitute this back into the original equation:

Multiply both sides by 2:

Use the logarithm property $n \log_b a = \log_b a^n$:

Since the bases are the same, the arguments must be equal:

Expand the right side:

Rearrange into a standard quadratic form $ax^2+bx+c=0$:

Factor the quadratic equation:

This gives two potential solutions:

Recall that the valid domain for $x$ is $x > 3$.
For $x=2$: This solution is not valid because $2 \ngtr 3$.
For $x=5$: This solution is valid because $5 > 3$.

There is only one valid solution, $x=5$.

The number of solutions is one.

The final answer is $\text{One}$.