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AhnSoHee
@AhnSoHee
December 2021
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Помогите решить алгебру
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sedinalana
Verified answer
74(2)
log(3)a=2,log(3)b=6
log(3)[a²/9b^1/4]=2log(3)a-log(3)9-1/4*log(3)b=2*2-2-1/4*6=4-2-1,5=0,5
75(2)
log(2)0,8-log(2)1 1/8+log(2)22,5=log(2)[4/5:9/8*45/25]=
=log(2)[4/5*8/9*45/2]=log(2)16=4
75(4)
5/3*log(2/3)8^1/5-3log(2/3)3+1/2log(2/3)36=
=log(2/3)∛8-log(2/3)3³+log(2/3)√36=log(2/3)[2*6/27)=log(2/3)(4/9)=2
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Answers & Comments
Verified answer
74(2)log(3)a=2,log(3)b=6
log(3)[a²/9b^1/4]=2log(3)a-log(3)9-1/4*log(3)b=2*2-2-1/4*6=4-2-1,5=0,5
75(2)
log(2)0,8-log(2)1 1/8+log(2)22,5=log(2)[4/5:9/8*45/25]=
=log(2)[4/5*8/9*45/2]=log(2)16=4
75(4)
5/3*log(2/3)8^1/5-3log(2/3)3+1/2log(2/3)36=
=log(2/3)∛8-log(2/3)3³+log(2/3)√36=log(2/3)[2*6/27)=log(2/3)(4/9)=2