Question

Log3 (2x-7) >= 2 log3(x+1) - log3(x-19)​

Answer 1

ОДЗ

2x-7>0; x>3.5

x+1>0;x>-1

x-19.0; x>19

общая ОДЗ x>19

log(3)(2x-7)-2log(3)(x+1)+log(3)(x-19)≥0

log(3)((2x-7)(x-19)/(x+1)^2))≥0

(2x-7)(x-19)/(x+1)^2≥1

(2x-7)(x-19)/(x+1)^2-1≥0

(2x^2-38x-7x+133-x^2-2x-1)/(x+1)^2≥0

(x^2-47x+132)/(x+1)^2≥0

(x-44)(x-3)/(x+1)^2≥0

----(-1)+++[3]----[44]+++

учитывая ОДЗ ответ x=[44;+∞)

Answer 2

To solve the inequality log3(2x-7) >= 2log3(x+1) - log3(x-19), we need to apply logarithmic properties to simplify the expression. Remember that we can use the following properties of logarithms:

1. log_a(b) >= log_a(c) if and only if b >= c (for a > 1)
2. log_a(b) + log_a(c) = log_a(b * c)
3. log_a(b) - log_a(c) = log_a(b / c)

Let's begin by simplifying the right-hand side:

2log3(x+1) - log3(x-19) = log3((x+1)^2) - log3(x-19) [Using the property log_a(b^m) = m * log_a(b)] = log3((x+1)^2 / (x-19)) [Using the property log_a(b) - log_a(c) = log_a(b / c)]

Now, the inequality becomes:

log3(2x-7) >= log3((x+1)^2 / (x-19))

Now we can remove the logarithm on both sides by exponentiating with base 3:

3^(log3(2x-7)) >= 3^(log3((x+1)^2 / (x-19)))

By the property 3^(log3(a)) = a, the inequality simplifies to:

2x - 7 >= (x+1)^2 / (x-19)

To solve this inequality, we need to find the critical points, where the expression on the right-hand side becomes zero or undefined. The critical points are when the denominator is equal to zero:

x - 19 = 0 x = 19

Now we have three intervals to test: (-∞, 19), (19, 7/2), and (7/2, ∞).

1. Test x in (-∞, 19): Choose x = 0: 2(0) - 7 >= (0+1)^2 / (0-19) -7 >= 1/19 (true)

2. Test x in (19, 7/2): Choose x = 5: 2(5) - 7 >= (5+1)^2 / (5-19) 3 >= -36/14 (false)

3. Test x in (7/2, ∞): Choose x = 10: 2(10) - 7 >= (10+1)^2 / (10-19) 13 >= 121/9 (true)

So, the solution to the inequality is x ∈ (-∞, 19] ∪ (7/2, ∞).