Question

2lg(-x)>lg(x+6) одз: х+6>0, х>-6

Answer 1

Одз:
x+6 > 0
-x > 0

x > -6
x < 0

2lg(-x) > lg(x+6)
lg(-x)^2 > lg(x+6)
x^2 > x + 6
x^2 - x - 6 > 0
x < -2
x > 3 (не принадлежит одз)

x < -2,
x > -6,
x < 0, откуда -6 < x < -2

Ответ: (-6; -2)

Answer 2

To solve the inequality 2lg(-x) > lg(x + 6), we need to consider the domain of the logarithmic functions. Remember that the logarithm is only defined for positive numbers. So, we have two conditions to satisfy:

1. -x > 0 (Since lg(-x) is defined only for -x > 0)
2. x + 6 > 0 (Since lg(x + 6) is defined only for x + 6 > 0)

Let's solve these conditions one by one:

1. -x > 0 Divide both sides by -1, and since we are dividing by a negative number, the inequality's direction will change: x < 0

2. x + 6 > 0 Subtract 6 from both sides: x > -6

Now, we have both conditions satisfied: x < 0 (from condition 1) x > -6 (from condition 2)

To find the combined solution, we need to find the intersection of these two inequalities. That is, we need to find values of x that satisfy both conditions simultaneously:

So, the solution is: -6 < x < 0.

In interval notation: (-6, 0).