lg(x2 – 2х) – lg(2x + 12) = 0

Предыдущий
вопрос Следующий
вопрос

To solve the equation lg(x^2 - 2x) - lg(2x + 12) = 0, we can use the properties of logarithms.

First, we know that log(a) - log(b) = log(a/b). Therefore, we can rewrite the equation as:

lg((x^2 - 2x)/(2x + 12)) = 0

Now, we can rewrite this equation in exponential form:

10^0 = (x^2 - 2x)/(2x + 12)

Simplifying 10^0 gives us:

1 = (x^2 - 2x)/(2x + 12)

Now, we can cross multiply to solve for x:

x^2 - 2x = 2x + 12

x^2 - 4x - 12 = 0

Now we have a quadratic equation that we can solve using the quadratic formula:

x = (-(-4) ± √((-4)^2 - 4(1)(-12))) / 2(1)
x = (4 ± √(16 + 48)) / 2
x = (4 ± √64) / 2
x = (4 ± 8) / 2

This gives us two possible solutions:

x = (4 + 8) / 2 = 12 / 2 = 6
x = (4 - 8) / 2 = -4 / 2 = -2

Therefore, the solutions to the equation lg(x^2 - 2x) - lg(2x + 12) = 0 are x = 6 and x = -2.