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log23______log32(填">"或"<")

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

(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,

[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

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

由对数函数的图象和性质可得:log23>1,log0.53<0而:4? 32=18∴log30.5<4?32<log32故答案为:log30.5<4?32<log32

(1)(2764)?13+(214)12+12log26=(6427)13+(94)12+12log26=43+32+16=3(2)(log23+log89)?(log34+log98+log32)=(log23+23log23)?(2log32+ 32log32+log32)=53log23?92log32=152

∵log0.32<log12=0,∴c<0.∵log13<log23<log33=1,∴0<a<1.∵log22<log32,∴1<b.综上可得:c<a<b.故答案为c<a<b.

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

∵a=log3π>1,b=log23=12log23<1,c=log32=12log32<1∴a>b,a>c.又log23>1>log32,∴b>c,∴a>b>c.故答案为 a>b>c

∵a=log23+log23=log233>log24=2,b=log233,c=log32<log33=1,∴a=b>c.故选:A.

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