Question

Tolong pls no 20 pakai cara ya

Answer 1

Verified answer

Jawaban : D

caranya ada di gambar
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Answer 2

Jawabannya:
(2x)^1+2log2x>64x³
=(2x)^1+2log2x+2logx>2^6x³
=(2x)^1+1+2logx>2^6x³
=(2x)²+2logx>2^6x³
=(2x)²(2x)2logx>2^6x³
=(2x)^2logx>2⁴x
=2^2logx.x^2logx>2⁴x
=x.x^2logx>2⁴x
=x^2logx>x^xlog2⁴
=2logx>xlog2⁴
=2logx-4^xlog2>0
=2logx-4/2logx>0
=2log²x-4>0

Misal y=2logx
maka:
y²-4>0
(y+2)(y-2)>0
y<-2 atau y>2

(i) y<-2
2logx<-2
2logx<2log2^-2
x<1/4

(ii) y>2
2logx>2
2logx>2log2²
x>4

HP={ x | x < 1/4 atau x > 4 }

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