log4(3)+log8(3) 怎么算

log4(3)+log8(3) 怎么算 log4(3)+log8(3) 怎么相加? log4(3)-log4(12)-log8(4) =log4(3/12)- log8( 8^(2/3) ) log8( 8^(2/3) )怎么得出来的? (log4(3)+log8(3))*log3(2) log3^36-log3^4+log2(log2^16)+8^2/3+(27/8)^-1/3+log8^27/log4^9怎么算啊,不好意思哦,我已经没分啦.麻烦一下写一写log2(log2^16) (27/8)^-1/3 log8^27/log4^9 (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) 怎么算的我知道（1/2）log2(3)+（1/3）log2(3) 但是下一步怎么算的↓（5/6）log2(3) [log4(3)+log8(3)]*log3(2)答案是1/3么