log8(9)乘以log3(32)-lg4-2lg5+3+27^(2/3)说明)-lg4-2lg5怎

发布时间:2021-02-19 21:45:28

log8(9)乘以log3(32)-lg4-2lg5+3+27^(2/3)说明)-lg4-2lg5怎么算

网友回答

-lg4-2lg5
=-(lg4+lg25)
=-[lg(4×25)]
=-lg100
=-2log8(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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