To solve this equation, we can first combine the logarithms on the left side using the product rule of logarithms:
lg(x-1)+lg(x-3) = lg((x-1)(x-3))
Now we have:
lg((x-1)(x-3)) = lg(3/2 x-3)
Now we can drop the logarithm on both sides and solve for x:
(x-1)(x-3) = 3/2 x - 3x^2 - 4x + 3 = 3/2 x - 32x^2 - 8x + 6 = 3x - 62x^2 - 11x + 12 = 0
Now we can factor the quadratic equation:
(2x-3)(x-4) = 0
Setting each factor to zero, we get:
2x - 3 = 0 or x - 4 = 0
Solving for x:
2x = 3 or x = 4
Therefore, the solutions to the equation lg(x-1)+lg(x-3) = lg(3/2 x-3) are x = 3/2 or x =4.
To solve this equation, we can first combine the logarithms on the left side using the product rule of logarithms:
lg(x-1)+lg(x-3) = lg((x-1)(x-3))
Now we have:
lg((x-1)(x-3)) = lg(3/2 x-3)
Now we can drop the logarithm on both sides and solve for x:
(x-1)(x-3) = 3/2 x - 3
x^2 - 4x + 3 = 3/2 x - 3
2x^2 - 8x + 6 = 3x - 6
2x^2 - 11x + 12 = 0
Now we can factor the quadratic equation:
(2x-3)(x-4) = 0
Setting each factor to zero, we get:
2x - 3 = 0 or x - 4 = 0
Solving for x:
2x = 3 or x = 4
Therefore, the solutions to the equation lg(x-1)+lg(x-3) = lg(3/2 x-3) are x = 3/2 or x =4.