First, we can combine the logarithms on the left side using the product rule for logarithms:
lgx + lg(x+1) = lg(x(x+1)) = lg(x^2 + x)
Now, we have:
lg(x^2 + x) = lg(2x^2 - 6)
Since the bases of the logarithms are the same, we can drop the logarithms and set the expressions inside them equal to each other:
x^2 + x = 2x^2 - 6
Rearranging the terms gives us a quadratic equation:
x^2 - x - 6 = 0
Now we can factor this quadratic equation:
(x-3)(x+2) = 0
Setting each factor equal to zero gives us the possible values of x:
x - 3 = 0 or x + 2 = 0
x = 3 or x = -2
Therefore, the solutions to the equation lgx + lg(x+1) = lg(2x^2-6) are x = 3 and x = -2.
First, we can combine the logarithms on the left side using the product rule for logarithms:
lgx + lg(x+1) = lg(x(x+1)) = lg(x^2 + x)
Now, we have:
lg(x^2 + x) = lg(2x^2 - 6)
Since the bases of the logarithms are the same, we can drop the logarithms and set the expressions inside them equal to each other:
x^2 + x = 2x^2 - 6
Rearranging the terms gives us a quadratic equation:
x^2 - x - 6 = 0
Now we can factor this quadratic equation:
(x-3)(x+2) = 0
Setting each factor equal to zero gives us the possible values of x:
x - 3 = 0 or x + 2 = 0
x = 3 or x = -2
Therefore, the solutions to the equation lgx + lg(x+1) = lg(2x^2-6) are x = 3 and x = -2.