Даю 20 б
Помогите пожалуйста срочно
с решение не могу разобраться
log(3)log(2)8
2log(27)log(10)1000
1/3log(9)log(2)8
log(4)log(16)256+log(4)корень из2
3log(2)log(4)16+log(1/2)2

log₃log₂8=log₃log₂2³=log₃(3*log₂2)=log₃3=1.2*log₂₇log₁₀1000=2*log₂₇log₁₀10³=2*log₃³(3*log₁₀10)=(2/3)*log₃3=2/3.(1/3)*log₉log₂8=(1/3)*log₉log₂2³=(1/3)*log₃²(3*log₂2)=(1/6)*log₃3=1/6.log₄log₁₆256+log₄√2= log₄log₄²4⁴+log₂²2⁻¹=log₂²((1/2)*log₄4⁴)+(1/2)*log₂2⁻¹==(1/2)*log₂2-(1/2)*log₂2=1/2-1/2=0.3*log₂log₄16+log₁/₂2=3*log₂log₄4²-log₂2=3*log₂(2*log₄4)-1=3log₂2-1=3-1=2.