To solve this logarithmic equation, we first need to isolate the logarithms on both sides:
lg(x^2-2x) = lg(2x+12)
Next, since the bases of the logarithms are the same (base 10), we can drop the logarithms and set the arguments equal to each other:
x^2 - 2x = 2x + 12
Now we have a quadratic equation. Let's bring all the terms to one side:
x^2 - 4x - 12 = 0
Now we can factor this quadratic equation:
(x - 6)(x + 2) = 0
Setting each factor to zero, we get:
x - 6 = 0 or x + 2 = 0
This gives us two possible solutions:
x = 6 or x = -2
Therefore, the solutions to the equation lg(x^2-2x) = lg(2x+12) are x = 6 and x = -2.
To solve this logarithmic equation, we first need to isolate the logarithms on both sides:
lg(x^2-2x) = lg(2x+12)
Next, since the bases of the logarithms are the same (base 10), we can drop the logarithms and set the arguments equal to each other:
x^2 - 2x = 2x + 12
Now we have a quadratic equation. Let's bring all the terms to one side:
x^2 - 4x - 12 = 0
Now we can factor this quadratic equation:
(x - 6)(x + 2) = 0
Setting each factor to zero, we get:
x - 6 = 0 or x + 2 = 0
This gives us two possible solutions:
x = 6 or x = -2
Therefore, the solutions to the equation lg(x^2-2x) = lg(2x+12) are x = 6 and x = -2.