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log23______log32(填">"或"<")

由于log23>log22=1,log32<log33=1,∴log23>log32,故答案为>.

[log3(2)+log2(3)]^2-log3(2)/log2(3)-log2(3)/log3(2) =[log3(2)+log2(3)]^2-[1/log2(3)]/log2(3)-[1/log3(2)]/log3(2) =[1/log2(3)+1/log3(2)]^2-[1/log2(3)]^2-[1/log3(2)]^2 =2/[log2(3)*log3(2)] =2

由对数函数的性质可知,log23>log22=1,log0.53<log0.51=0,由指数函数的性质可知,0<4?32=2-3<2?52<20=1,故log0.53<4?32<2?52<log23.故答案为:log0.53<4?32<2?52<log23.

=lg(20÷2)-(lg3/lg2)(lg2/lg3)+2(lg1/4÷lg2) =lg10-1+2(-2lg2÷lg2) =1-1-4 =-4

(lg5)^2 + lg2×lg50 + log 2 32 =(lg5)^2 + lg2×lg(5*10) + log 2 (2^5) =(lg5)^2 + lg2×(lg5 +lg10) + 5 =(lg5)^2 + lg2×(lg5 +1) + 5 =(lg5)^2 + lg2×lg5 +lg2 + 5 = lg5 * ( lg2+lg5) +lg2 + 5 = lg5 * lg(2*5) +lg2 + 5 = lg5 * l...

(1)log23?log32=log23?log22log23=log22=1;(2)(3?π)2=|3-π|,因为3-π<0,所以(3?π)2=|3-π|=π-3,故答案为:1,π-3,

(1)由于函数y=log2x 在(0,+∞)上是增函数,且2<3,故有 log22<log23,故答案为<.(2)由于log32<log33=1,故答案为<.(3)由于log134<log131=0,故答案为<.

∵1=log33>a=log32>log31=0,b=log52<log32=a,c=log23>log22=1,∴c>a>b.故答案为:c>a>b.

(log23+log83)(log32+log92)=(log833+log83)(log94+log92)=log893×log98=lg93lg8×lg8lg9=lg352lg32=522=54.故选:A.

(log23+log43)(log32+log92)=(log23+log23)(log32+log32)=log2(33)log3(22)=32log23?32log32=94.故答案为:94.

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