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

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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.**