Question

Log4(16-16x) < log4 (x^2-3x +2)+log4 (x+6)

Answer 1

Решение приложено

=============================================================

Answer 2

Ответ:

x∈∅

Объяснение:

log₄ (16-16·x) < log₄ (x²-3·x+2)+log₄ (x+6)

ОДЗ (область допустимых значений):

16-16·x>0, x²-3·x+2>0, x+6>0 ⇔ 1>x, (x-1)·(x-2)>0, x>-6 ⇔

⇔ x∈(-∞; 1), x∈(-∞; 1)∪(2; +∞), x∈(-6; +∞) ⇔ x∈(-6; 1).

Решение.

log₄ (16-16·x) < log₄ (x²-3·x+2)·(x+6), так как 4>1 :

(16-16·x) < (x²-3·x+2)·(x+6)

0<(x-1)·(x-2)·(x+6)-16·(1-x)

(x-1)·(x-2)·(x+6)+16·(x-1)>0

(x-1)·((x-2)·(x+6)+16)>0

(x-1)·(x²+4·x-12+16)>0

(x-1)·(x²+4·x+4)>0

(x-1)·(x+2)²>0, так как строгое неравенство, то x≠-2, тогда

x-1>0

x>1

x∈(1; +∞).

Вместе с ОДЗ:

x∈(1; +∞)∩(-6; 1) ⇒ x∈∅.

Answer 3

To solve the inequality log₄(16-16x) < log₄(x²-3x+2) + log₄(x+6), we can start by simplifying the equation using logarithmic properties.

1. Combine the logarithms on the right side using the product rule of logarithms: log₄(16-16x) < log₄((x²-3x+2)(x+6))

2. Apply the power rule of logarithms to the right side: log₄(16-16x) < log₄((x-1)(x-2)(x+6))

3. Now, we can eliminate the logarithms by exponentiating both sides with base 4: 16-16x < (x-1)(x-2)(x+6)

4. Expand the right side: 16-16x < (x³-3x²+2x-2x²+6x-12)(x+6)

5. Simplify further: 16-16x < (x³-5x²+8x-12)(x+6)

6. Expand the right side again: 16-16x < x⁴+x³-5x³-5x²+8x²-40x+8x-12x²+60-12

7. Combine like terms: 16-16x < x⁴-4x³-9x²-32x+56

8. Move all terms to the left side to get a quadratic equation: x⁴-4x³-9x²-32x+56 - 16 + 16x < 0

9. Simplify: x⁴-4x³-9x²-16x+40 < 0

Now, we have a quartic inequality that we can solve. However, solving quartic equations can be complex and cumbersome. It is challenging to find an exact solution algebraically in most cases. You might consider using numerical methods or graphing the equation to determine the range of values for which the inequality holds true.