解该式：(log4(3)+log8(3))*(lg2/lg3)log4(3)是指以4为底

解该式：(log4(3)+log8(3))*(lg2/lg3)log4(3)是指以4为底 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? 化简此式（log4 3+log8 3)×（log3 2+log9 2） (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） 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!