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

logx+log(x-3) = 1.

log(x(x-3) = 1, as loga+logb = log(ab). Taking antologarithms with respect to base 10, we get:

x(x-3) = 10^1.

x^x-3x -10 = 0. Or

x = [3+ srt(3^2+4*10)}/2 = (3+sqrt49)/2 = 5. Or

x= (3-7)/2 = -2

For the beginning, we'll use the product property of logarithms: the sum of logarithms is the logarithm of the product.

lgx+lg(x-3)=1

lg [x(x-3)] = 1

The base of logarithm being 10, we'll re-write the equation:

[x(x-3)] = 10^1

We'll open the brackets:

x^2 - 3x = 10

We'll add -10 both sides of the equation:

x^2 - 3x - 10 = 0

We'll solve the quadratic equation:

x1 = [3 + sqrt(9+40)]/2

x1 = (3+7)/2

x1 = 5

x2 = (3-7)/2

x2 = -4/2

x2 = -2

The second solution, namely -2, is not convenient, because lg(-2) is undefined.

We'll check the first solution into equation:

lg5+lg(5-3)=1

lg5 + lg2 = 1

lg(5*2) = 1

lg 10 = 1

10 = 10^1

10 = 10

So, the solution x=5 is the only solution of the equation.