lgx+ lg(x-3)=1
We know that:
logx+lgy= lgxy
and, log10=1
Now substitute;
lgx(x-3)=lg10
==> x(x-3)=10
==> x^2-3x = 10
==> x^2 -3x -10=0
Now factorize:
==> (x-5)(x+2)=0
x1= 5
x2=-2 ( this solution is impossible because log-2 is undefined and x can not be a negative value)
Then the only solution is x1=5
To check:
lg5 + lg(5-3)= lg5+lg2 = lg10=1