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Marjangul1
@Marjangul1
August 2022
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log5(3x+1)>log5(x-2)
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kirichekov
Verified answer
Log₅(3x+1)>log₅(x-2)
ОДЗ:
x>2. x∈(2;∞)
основание логарифма а=5, 5>1. знак неравенства не меняем:
3x+1>x-2
2x>-3
x>-1,5
учитывая ОДЗ, получим
=> x>2
x∈(2;∞)
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Verified answer
Log₅(3x+1)>log₅(x-2)ОДЗ:
x>2. x∈(2;∞)
основание логарифма а=5, 5>1. знак неравенства не меняем:
3x+1>x-2
2x>-3
x>-1,5
учитывая ОДЗ, получим
=> x>2
x∈(2;∞)