各位大神log4(3)log9(2)-log1/2(4√32)怎做啊?是不是？

各位大神log4(3)log9(2)-log1/2(4√32)怎做啊?是不是？ 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 2)*(log4 3+log8 3)求值 化简(log4(3)+log8(3))(log3(2)+log9(2)) (log3^2 + log9^4)×(log4^3 + log8^3) 计算log9(根号2)*[log4(3)+log8(3)] 计算:(log4 3+log8 3)×log9 √2=? 求值:(log4^ 3+log8 ^9)*(log3^ 2+log9 ^16) log9×log4 (log4(3)+log8(3))(log3(2)+log9(2))-log4(3)×log1/3(四次根号32) 2log4 (2-根号3)+3log9(2+根号3)错了 是2^(log4 (2-根号3))+3^(log9(2+根号3)) 急 求 （log3 2+log9 2）-（log4 3+log8 3）求 （log3 2+log9 2）-（log4 3+log8 3） （1）log4 27乘以log25 8乘以log9 5 （2）（log4 3+log8 3）（log3 2+log9 2）、 2^log4(2-√3) ^2+ 3^log9(2+√3)^2如题 2^log4(√3+2)^2+3^log9(√3-2)^2= 2^(log4 (2-根号3))+3^(log9(2+根号3))= 2log4 (2-根号3)²+3log9(2+根号3)²