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Whithoutme
@Whithoutme
July 2022
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модуль x-13 модуль*log в основании 2 аргумент (x-3) = 3*(13-x)
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Dимасuk
Verified answer
2 votes
Thanks 2
sedinalana
|x-13|*log(2)(x-3)=3*(13-x)
ОДЗ
x-3>0⇒x∈(3;∞)
1)3<x<13
(13-x)*log(2)(x-3)=3(13-x)
log(2)(x-3)=3
x-3=8
x=8+3
x=11
2)x=13
0*log(2)10=3*0
0=0
3)x>3
(x-13)*log(2)(x-3)=-3*(x-13)
log(2)(x-3)=-3
x-3=1/8
x=3+1/8
x=3 1/8 не удов усл
Ответ x=13,x=11
1 votes
Thanks 2
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Answers & Comments
Verified answer
ОДЗ
x-3>0⇒x∈(3;∞)
1)3<x<13
(13-x)*log(2)(x-3)=3(13-x)
log(2)(x-3)=3
x-3=8
x=8+3
x=11
2)x=13
0*log(2)10=3*0
0=0
3)x>3
(x-13)*log(2)(x-3)=-3*(x-13)
log(2)(x-3)=-3
x-3=1/8
x=3+1/8
x=3 1/8 не удов усл
Ответ x=13,x=11