题目
[(1-log6 3)^2+log6 2-log6 18]/log6 4
提问时间：2020-08-07
答案
[(1-log6 3)^2+log6 2-log6 18]/log6 4
=[(log6 6-log6 3)^2+log6 2-log6 (2*9)]/log6 (2^2)
=[(log6 6/3)^2+log6 2-(log6 2 +log6 9)]/(2*log6 2)
=[(log6 2)^2+log6 2-log6 2-log6 9]/(2*log6 2)
=[(log6 2)^2-log6 9]/(2*log6 2)
=(log6 2)/2-log6 3/log6 2
=(log6 2)/2-log2 3
=[(log6 6-log6 3)^2+log6 2-log6 (2*9)]/log6 (2^2)
=[(log6 6/3)^2+log6 2-(log6 2 +log6 9)]/(2*log6 2)
=[(log6 2)^2+log6 2-log6 2-log6 9]/(2*log6 2)
=[(log6 2)^2-log6 9]/(2*log6 2)
=(log6 2)/2-log6 3/log6 2
=(log6 2)/2-log2 3
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