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Task/26406750
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Log₂ (2ˣ -1) *Log₂ 2(2ˣ -1) /(-1) > - 2 ;
Log₂ (2ˣ -1) *Log₂ 2(2ˣ -1) < 2 ;
Log₂ (2ˣ -1) *(Log₂ (2ˣ -1) +1 ) < 2 ; замена: t = Log₂ (2ˣ -1)
t² +t -2 < 0 ;
(t +2)(t -1) <0 ⇒ -2 < t <1;
- 2 < Log₂ (2ˣ -1) < 1 ;
2⁻² < 2ˣ -1 < 2¹ ;      * * * 2 >1* * *
1/4 +1 <  2ˣ < 2 +1 ;
5/4 < 2ˣ < 3 ;
Log₂ 5/4 <x < Log₂ 3 .    * * * 2 >1* * *

ответ : x ∈ ( Log₂ 5/4 ; Log₂ 3) .

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MizoriesKun ㏒₂(2ˣ-1)* ㏒₁/₂(2ˣ⁺¹- 2) >-2

㏒₂(2ˣ-1)*(-1* ㏒₂(2*(2ˣ -1) > -2

㏒₂(2ˣ-1)*(-1*( ㏒₂2 +㏒₂(2ˣ -1) > -2

㏒₂(2ˣ-1)*(-㏒₂2 - ㏒₂(2ˣ -1) > -2 ОДЗ 2ˣ-1>0 2ˣ>1 x>0

㏒₂(2ˣ-1)*( -1 - ㏒₂(2ˣ -1) > -2 замена ㏒₂(2ˣ-1)=а

а*(-1-а) > -2

-а-а²> -2

а²+а-2<0

D=1+8=9

a=(-1+3)/2=1 ㏒₂(2ˣ-1)=1 (2ˣ-1)=2 2ˣ =3 x=㏒₂ 3

a=(-1-3)/2=-2 ㏒₂(2ˣ-1)=-2 (2ˣ-1)=1/4 2ˣ=5/4 x=㏒₂5/4

+ - +
-------㏒₂5/4---------- ㏒₂ 3-----------

x∈(㏒₂5/4 ; ㏒₂ 3)

1 votes Thanks 0

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