log4 sin(3π)/4+log9 tan[(-5π)/6]=_______.

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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
再问： √2/2怎么是化成2^(1/2)?2^(1/2)不是√2吗？
再答：你说得对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 =0
再问：不是应该tan(7π/6)=tan30°=√3/3么
再答：怎么做题，不细心,不好意思 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 （√3/3） =log4 (4^(-1/4))+log9 9^(-1/4) =-1/4-1/4 =-1/2

1.log2 3*log3 4*log4 5*log5 2 2.(log4 3+log8 3)8(log3 2+log9 (log4 3+log8 3)(log3 5+log9 5)(log5 2+log25 2)= 若log2^3×log3^4×log4^x=log9^3 ,求x （log2 3+log4 27）*（log3 4+log9 8）= (log3^2 + log9^4)×(log4^3 + log8^3) 计算:(log4 3+log8 3)×log9 √2=? log2 根号2+log9 27+4^log4 13=? [log2(3)+log4(9)][log3(4)+log9(2)]等于计算:(1/4)^log4^1/3-lg25-lg4-log9^27 log4(3)*log9(2)+log2(4倍根号下64) log9(32)*log64(2 (log9(8))*(log4(3根号3)）? log4^27×log(8)25×log9^5