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log8(9)乘以log3(32)-lg4-2lg5+3+27^(2/3)
说明)-lg4-2lg5怎么算
人气:452 ℃ 时间:2020-02-06 00:50:07
解答
-lg4-2lg5
=-(lg4+lg25)
=-[lg(4×25)]
=-lg100
=-2
log8(9)乘以log3(32)-lg4-2lg5+3+27^(2/3)
=(2lg3)/(3lg2) × (5lg2)/(lg3) - (lg4+lg25) +3 +9
=10/3 - 2+12
=40/3
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