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IR72016
@IR72016
July 2022
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7/2*log2(x) + log4(x)=4
Не могу никак решить
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kirichekov
Verified answer
Log₄x=log_(₂²) x=(1:2)*log₂x=(1/2)*log₂x
(7/2)*log₂x+(1/2)*log₂x=4
ОДЗ: x>0
(7/2+1/2)*log₂x=4
4*log₂x=4
log₂x=1
x=2¹
x=2
3 votes
Thanks 4
IR72016
В задании сумма корней уравнения... видимо опечатка
kirichekov
данном задании один корень
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Answers & Comments
Verified answer
Log₄x=log_(₂²) x=(1:2)*log₂x=(1/2)*log₂x(7/2)*log₂x+(1/2)*log₂x=4
ОДЗ: x>0
(7/2+1/2)*log₂x=4
4*log₂x=4
log₂x=1
x=2¹
x=2