Solve for real x: lg(x −1) + lg(6x − 5) = 2

We'll impose the constraints of existence of
logarithms:

x - 1 > 0 => x >
1

6x - 5 > 0 => 6x > 5 => x
> 5/6

The common interval of admissible values that
makes the logarithms to exist is (1 ; +infinite).

Since the
bases of logarithms are matching, we'll apply the product
rule:

lg(x −1) + lg(6x − 5) = lg
(x-1)(6x-5)

We'll create matching bases both sides,
therefore we'll write 2 = lg 100

We'll re-write the
equation:

lg (x-1)(6x-5) = lg
100

Since the bases of logarithms are matching, we'll apply
one to one rule:

(x-1)(6x-5) =
100

We'll remove the
brackets:

6x^2 - 5x - 6x + 5 =
100

6x^2 - 11x + 5 - 100 =
0

6x^2 - 11x - 95 = 0

We'll
apply the quadratic formula:

x1 = [11 + sqrt(2401
)]/12

x1 = (11 +49 )/12

x1 =
5

x2 = -38/12

x2 =
-19/6

Since the 2nd value of x does not
belong to the interval (1 ; +infinite), we'll keep as solution only the positive value
of x: x = 5.