Equations & Inequations

$\begin{aligned} \log_{x-3} \left(x^3-3x^2-4x+8 \right) & = 3 \\ \implies x^3-3x^2-4x+8 & = (x-3)^3 \\ x^3-3x^2-4x+8 & = x^3 - 9x^2 + 27x - 27 \\ 6x^2 - 31x + 35 & = 0 \end{aligned}$

$\implies x = \dfrac {31 \pm \sqrt{31^2-4(6)(35}}{12} = \dfrac 72, \ \dfrac 53$ .

Note that the base of logarithm $\log_{\color{#D61F06}x-3}$ must be $> 0$ , that is, ${\color{#D61F06} x-3} > 0$ $\implies x > 3$ then only $x = \dfrac 72$ is a solution. Therefore, the answer is $\boxed{1}$ .