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Pashka1993
@Pashka1993
July 2022
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Помогите найти множество решений неравенства:
log2/7(2x-28) > log2/7(6x)
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nKrynka
Решение
log2/7(2x-28) > log2/7(6x)
ОДЗ: 2х - 28 > 0; x > 14; 6x> 0, x > 0. Отсюда х > 14
так как 0
< 2/7 < 1, то
2x - 28 < 6x
2x - 6x < 28
- 4x < 28
x > - 7
С учётом ОДЗ
x > 14
Ответ: x > 14 или (14: + ≈)
2 votes
Thanks 1
Pashka1993
2х - 28 > 0; x > 12 разве х не 14?
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Answers & Comments
log2/7(2x-28) > log2/7(6x)
ОДЗ: 2х - 28 > 0; x > 14; 6x> 0, x > 0. Отсюда х > 14
так как 0 < 2/7 < 1, то
2x - 28 < 6x
2x - 6x < 28
- 4x < 28
x > - 7
С учётом ОДЗ x > 14
Ответ: x > 14 или (14: + ≈)