Question

5 log7 (x^2 - 7x +12) меньше либо равно 6+ log7 (x-4)^5/(x-3)

Answer 1

5log₇(x² - 7x + 12) ≤ 6 + log₇[(x - 4)⁵ /(x - 3)]
Разложим на множители x² - 7x + 12
x² - 7x + 12 = 0
D = 49 - 48 = 1
x1 = 0.5(7 - 1) = 3    x2 = 0.5(7 + 1) = 4
(x² - 7x + 12) = (x - 3)(x - 4)
log₇[(x - 3)⁵(x - 4)⁵ ≤ 6 + log₇[(x - 4)⁵ /(x - 3)]
ОДЗ
[(x - 3)(x - 4)]⁵ > 0 
[(x- 4)⁵/(x - 3) > 0
методом интервалов получаем ОДЗ
x∈ (-∞; 3) U (4; +∞)
log₇[(x - 3)⁵(x - 4)⁵ -  log₇[(x - 4)⁵ /(x - 3)] ≤ 6
log₇[(x - 3)⁵(x - 4)⁵(x - 3)/(x - 4)⁵] ≤ 6
x ≠ 4
сокращаем на (x - 4)⁵
log₇(x - 3)⁶ ≤ 6
log₇(x - 3)⁶ ≤ log₇7⁶
(х - 3)⁶ ≤ 7⁶
Если х - 3 > 0, то х > 3 и
х - 3 ≤ 7    ----->    x ≤ 10, тогда х ∈ (3; 10]
Если х - 3 < 0, то х < 3 и х ≤ 10, тогда  х ∈ (-∞; 3)
Сопоставляя полученное решение с ОДЗ, запишем ответ:
x∈ (-∞; 3) U (4; 10]

Answer 2

To solve this inequality, we can start by simplifying both sides using logarithm rules. Remember that the logarithm of a product is the sum of the logarithms of the factors, and the logarithm of a quotient is the difference of the logarithms of the numerator and denominator. Applying these rules, we get:

5 log7 (x^2 - 7x +12) <= 6+ log7 [(x-4)^5/(x-3)]

Next, we can use the rule that says we can move the logarithm inside the argument as an exponent. This gives us:

log7 [(x^2 - 7x +12)^5] <= log7 [7^6 * (x-4)^5/(x-3)]

Now we can remove the logarithm on both sides by taking the exponential of base 7. This gives us:

(x^2 - 7x +12)^5 <= 7^6 * (x-4)^5/(x-3)

To simplify this expression, we can take the fifth root of both sides, which gives us:

x^2 - 7x +12 <= 7 * (x-4)/(x-3)

Expanding the right-hand side and simplifying, we get:

x^2 - 7x +12 <= 7x - 28

Moving all the terms to one side, we get:

x^2 - 14x + 40 <= 0

Factorizing the left-hand side, we get:

(x - 10)(x - 4) <= 0

The solution to this inequality is x <= 4 or x >= 10. However, we also need to check the denominator of the logarithm to make sure it is positive. For x <= 4, the denominator (x - 3) is negative, so this solution is not valid. Therefore, the only solution that satisfies both the inequality and the condition on the denominator is x >= 10.