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log1/3(1/2)=log3(2) 对吗 为什么

log1/2(2/3) =log1/2(2)-log1/2(3) =-1+lg2(3) =-1+1.585 =0.585

=lg0.5/lg3=-0.301/0.477=-0.631

(1/2)^[log2(3)] =[2^(-1)]^[log2(3)] =2^[(-1)*log2(3)] =2^[-log2(3)] =2^[log2(3)^(-1)] =2^[log2(1/3)]

解: a=log1/3[2]=-log3[2]>-log3[3]=-1,且ab

换底公式怎么写知道吧?将底换成1/3,左边分母就成了log1/3(1/3)/log1/3(1/2)

a=log(1/2) =log(1/2) /log(1/3) =log2 b=log(2/3) = log (2/3) /log(1/2) = (1- log2)/log2 =-1+ 1/log2 c= log (4/3) =-1+ log4 =-1+ 2log2 c- a = -1+ log2 a> c log 2 >0.5 c-b =2log2 - 1/log2 >0 => c>b ie a>c>b

=1/2的(log2[3]) =1/3 这里用到a^loga(N)=N

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