(log4(3)+log8(3))*(log3(2)+log8(2))-log1/2(^4√32)=?..
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(log4(3)+log8(3))*(log3(2)+log8(2))-log1/2(^4√32)=?..
(log4(3)+log8(3))*(log3(2)+log8(2))-log1/2(^4√32)=?
..
(log4(3)+log8(3))*(log3(2)+log8(2))-log1/2(^4√32)=?..
=(log(2,3)/2+log(2,3)/3)*(log(3,2)+1/3-?)
log(a,b)=ln(b)/ln(a)
log4(3)+log8(3) 怎么算
log4(3)+log8(3) 怎么相加?
(log4(3)+log8(3))*log3(2)
log4(3)-log4(12)-log8(4) =log4(3/12)- log8( 8^(2/3) ) log8( 8^(2/3) )怎么得出来的?
(log3^2 + log9^4)×(log4^3 + log8^3)
(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=?
求值:(log4^ 3+log8 ^9)*(log3^ 2+log9 ^16)
(log4(3)+log8(3))*(log3(2)+log8(2))-log1/2(^4√32)=?..
(log4(3)+log8(3))(log3(2)+log9(2))-log4(3)×log1/3(四次根号32)
急 求 (log3 2+log9 2)-(log4 3+log8 3)求 (log3 2+log9 2)-(log4 3+log8 3)
解该式:(log4(3)+log8(3))*(lg2/lg3)log4(3)是指以4为底
log3 4*log4 8*log8 m=log4 16,则m为[ ]其中3、4、8、4为底数.
[log4(3)+log8(3)]*log3(2)答案是1/3么
求(log3^2 + log9^2)×(log4^3 + log8^3)的值.请给出过程,谢谢!
求值:(log4(3)+log8(3))(log3(2)+log3(4))-log2(64)