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