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Warden11
@Warden11
July 2022
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Прошу помощи! Решите уравнение: [tex]( log_{2} (4x) )^{2} + log_{2} \frac{ x^{2} }{8} =8[/tex]
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Gviona
(log2(4x))^2+log2(x^2/8) =8
ОДЗ: x>0
(log2(4)+log2(x))(log2(4)+log2(x))+ log2(x^2)-log2(8)=8
(2+log2(x))*(2+log2(x))+2log2(x)-3=8
4+2log2(x)+2log2(x)+(log2(x))^2+2log2(x)=11
(log2(x))^2+6log2(x)-7=0
Замена: log2(x)=t
t^2+6t-7=0
D=6^2-4*1*(-7)=64
t1=(-6-8)/2=-7
t2=(-6+8)/2=1
Обратная замена:
1)log2(x)=-7
x=2^-7
x=1/128
2)log2(x)=1
x=2
Ответ: 1/128; 2
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Answers & Comments
ОДЗ: x>0
(log2(4)+log2(x))(log2(4)+log2(x))+ log2(x^2)-log2(8)=8
(2+log2(x))*(2+log2(x))+2log2(x)-3=8
4+2log2(x)+2log2(x)+(log2(x))^2+2log2(x)=11
(log2(x))^2+6log2(x)-7=0
Замена: log2(x)=t
t^2+6t-7=0
D=6^2-4*1*(-7)=64
t1=(-6-8)/2=-7
t2=(-6+8)/2=1
Обратная замена:
1)log2(x)=-7
x=2^-7
x=1/128
2)log2(x)=1
x=2
Ответ: 1/128; 2