Consider the equation $$|x-3|^{3x^2-10x+3}=1$$
Rewriting it in logarithmic form, $$\log_{|x-3|}1=3x^2-10x+3.$$
Since the base of a logarithm must be neither $0$ nor $1,$ $$3x^2-10x+3=0\\x=3\quad\text{or}\quad\frac13\\x=\frac13.$$
However, the given solution is $$x=\frac13\quad\text{or}\quad2\quad\text{or}\quad4.$$
Please tell me what I am missing here.
By definition (the expression $\log_ax$ carries implicit conditions), $$a^y=x\quad\textbf{and}\quad a\in(0,1)\cup(1,\infty) \iff y=\log_ax.\tag1$$
$$|x-3|^{3x^2-10x+3}=1.$$ Rewriting it in logarithmic form, $$\log_{|x-3|}1=3x^2-10x+3.$$
By the reverse direction of definition $(1),$ this step has restricted the solution set, so is invalid. Instead, $$|x-3|^{3x^2-10x+3}=1 \\\implies\\ \log_{|x-3|}1=3x^2-10x+3 \quad\text{or}\quad |x-3|=1\quad\text{or}\quad |x-3|=0.$$
The first disjunct gives the solution $\dfrac13$ (as you've shown), the second disjunct gives the remaining solutions $2$ and $4,$ while the third disjunct gives the extraneous solution $3.$
Actually, the logarithmic form is an unnecessary distraction in this exercise: \begin{align}{}&|x-3|^{3x^2-10x+3}=1\\\iff{}&|x-3|=1 \quad\text{or}\quad \big(x-3\ne0 \quad\text{and}\quad 3x^2-10x+3=0\big)\\\iff {}&x=\frac13\quad\text{or}\quad2\quad\text{or}\quad4.\end{align}
