Question 1140920
Probably there is a typo, and something got "lost in translation".
 
If the function is {{{f(x)=A*log((x+B))}}} or {{{y=A*log((x+B))}}} ,
the domain of the function is where {{{x+B>0}}} .
 
For point (2,0) where {{{x=2}}} and {{{y=0}}} ,
{{{0=A*log((2+B))}}}
Either {{{A=0}}}, or {{{log((2+B))=0}}}
The solution {{{A=0}}} would make the function {{{y=0}}} ,
even for {{{x=0}}} , so it must be
{{{log((2+B))=0}}} , which means 
{{{2+B=1}}} ,
{{{B=1-2}}} , and
{{{highlight(B=-1)}}}
Then, the function is only defined where {{{x-1>0}}}, or {{{x>1}}}
 
The point (0,3) with {{{x=0}}} cannot be part of the graph of such a function.