已知log2(6)=a 求log2(18)
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已知log2(6)=a 求log2(18)
∵log2(6)=log2(2*3)=log2(2)+log2(3)=1+log2(3)=a
∴log2(3)=a-1
∴log2(18)=log2(3*6)=log2(3)+log2(6)=a-1+a=2a-1
log(2)[6]=log(2)[2]+lg(2)[3]=a,则:
log(2)[3]=a-1
log(2)[18]
=log(2)[2]+log(2)[3²]
=1+2log(2)[3]
=1+2(a-1)
=2a-1
log2(18)
=log2(36)-log2(2)
=2log2(6)-1
=2a-1
2a-1
log2(6)=a
log2 2×3=a
log2 2+log2 3=a
∴1+log2 3=a
∴log2 3=a-1
log2 18=log2 2+2log2 3=1+2(a-1)=2a-1
log2(6)=log2(3)+log2(2)=a
log2(3)=a-1
log2(18)=log2(6)+log2(3)=a+a-1=2a-1
等于2a-1 ,因为log2(6)=log2(2x3)=log2(2)+log2(3)=1+log2(3)=a,所以log2(3)=a-1, 又log2(18)=log2(6x3)=log2(6)+log2(3)=a+(a-1)=2a-1
log2(6)=log2(2)+log2(3)=a,
所以log2(3)=a-1
log2(18)=log2(2)+log2(9)
=1+log2(3的平方)
=1+2log2(3)
=1+2(a-1)
=2a-1
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