Question 950668
I assume you have log base {{{10}}}
{{{y=2log(10,(x-3))}}}

first,  the vertical asymptote of y =2log(x-3) is at {{{x = 3}}}

you also need to remember that logs are not defined for {{{negative}}} {{{x}}} or for {{{x = 0}}}, 

so, the vertical asymptote is at {{{x = 3}}}, you just start with {{{x = 4}}} because {{{x=3}}} will give you {{{log(0)}}} which is {{{y = -infinity}}}, and work from there, using the definition of the log, make a table

{{{x}}}|....{{{y=2log(10,(x-3))}}}
{{{4}}}|....{{{0}}}.......{{{y=2log(10,(4-3))=0}}}
{{{5}}}|....{{{0.6}}}...{{{y=2log(10,(5-3))=0.6}}}
{{{6}}}|....{{{0.95}}}.....{{{y=2log(10,(6-3))=0.95}}}
{{{7}}}|....{{{1.2}}}.......{{{y=2log(10,(7-3))=1.2}}}

plot these points and draw a line through

{{{drawing( 600, 600, -5, 10, -5, 5,
line(3,5,3,-5),
circle(4,0,.12),circle(5,0.6,.12),locate(4,0,p(4,0)),locate(5,0.6,p(5,0.6)),
circle(6,0.95,.12),circle(7,1.2,.12),locate(6,0.95,p(6,0.95)),locate(7,1.2,p(7,1.2)),
 graph( 600, 600, -5, 10, -5, 5, 2*log(10,(x-3)) )) }}}