log4 sin(3π)/4+log9 tan[(-5π)/6]=_______.
来源:学生作业帮助网 编辑:作业帮 时间:2024/11/15 23:56:48
x)O7Q(0>ߠo
[*$Ek蚂Dbm!@&H$v6MmlJ5u27D4?߀(NY$k
,m ql@A QEpy
log4 sin(3π)/4+log9 tan[(-5π)/6]=_______.
log4 sin(3π)/4+log9 tan[(-5π)/6]=_______.
log4 sin(3π)/4+log9 tan[(-5π)/6]=_______.
log4 sin(3π)/4+log9 tan[(-5π)/6]
=log4 sin(π)/4+log9 tan[2π+(-5π)/6]
=log4 (√2/2)+log9 tan[7π/6]
=log4 (2^(1/2))+log9 √3
=log4 (4^(1/4))+log9 9^(1/4)
=1/4+1/4
=1/2
log4 sin(3π)/4+log9 tan[(-5π)/6]=_______.
计算log4(3)*log9(32)
log4(3)*log9(2)+log2(4倍根号下64) log9(32)*log64(2log4(3)*log9(2)+log2(4倍根号下64) log9(32)*log64(27)+log9(2)*log4(根27)
(log3^2 + log9^4)×(log4^3 + log8^3)
计算:(1/4)^log4^1/3-lg25-lg4-log9^27
log9×log4
log4(sin3π/4)+log9[tan(-5π/6)]的值是多少
如题{log(2√2) - 3log√3 - 3} / {log9 - log4 + 4}
(log3,4+log9,4)log4,3—log2,2根号2 这怎么解答了
若log2^3×log3^4×log4^x=log9^3 ,求x
对数函数计算 log4 3·log9 2-log1/2 4次根号32
各位大神log4(3)log9(2)-log1/2(4√32)怎做啊?是不是?
log√3^9+log9^27+(1/4)log4^1/16
log√3^9+log9^27+(1/4)log4^1/16
(log3 2+log9 2)*(log4 3+log8 3)求值
计算log9(根号2)*[log4(3)+log8(3)]
化简(log4(3)+log8(3))(log3(2)+log9(2))
计算:(log4 3+log8 3)×log9 √2=?