log8(9)乘以log3(32)-lg4-2lg5+3+27^(2/3)

来源：学生作业帮编辑：搜搜考试网作业帮分类：数学作业时间：2024/06/04 15:23:31

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

-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

log8(9)乘以log3(32)-lg4-2lg5+3+27^(2/3) log3(2)*log8(9) log3 (4√27)/3 +lg25 +lg4 +7^(log7 2) log3 5=m ,log8 3=n 求lg5 (log4(3)+log8(3))*log3(2) log3(2)*log5(3)*log8(5) lg5/2+lg4=? 求值:(log4^ 3+log8 ^9)*(log3^ 2+log9 ^16) (log4(3)+log8(3))*(log3(2)+log8(2))-log1/2(^4√32)=? [(lg4-lg60)/(lg3+lg5)]^3-4^5*2^-11 (log2(3)+log8(9))x(log3(4)+log9(8)+log3(2)) (log2 3+log8 9)(log3 4+log9 8+log3 2)的运算过程