Answer:
The answer is "[tex]\log(x-3)- \log(2)[/tex]"
Step-by-step explanation:
The asymptote is that when the group is 0 within. Let's all assume that log base 2 is: Log(ax-b); a position mostly online also they know. Let's use this (5,0) and log(5a-b)=0; they get out of it 5a-b=1; now that the results for a and b must be known. So we can say unless the asymptote is x=3, 3a-b=0; then we have the equations system, and with a and b we can solve
[tex]\to b=5a-1 \\\\ \therefore\\\\\to 3a-(5a-1)=0\\\\\to 3a-5a+1=0\\\\\to -2a+1=0\\\\\to -2a=-1\\\\\to a=\frac{1}{2} \ and\ b= \frac{3}{2}[/tex]
[tex]\to \log(\frac{x}{2}-\frac{3}{2})\\\\\to \log(\frac{(x-3)}{2})\\\\\to \log(x-3)- \log(2)[/tex]
