log以2为底2000
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log2(25)*log3(1/16)*log5(1/9)=[2log2(5)]*[-4log3(2)]*[-2log5(3)=[2*lg5/lg2]*[-4*lg2/lg3]*[-2*lg3/lg5
log816=log216/log28=4/3log23*log32=log23/log23=1结果为4/3+1=7/3
再问:第二步到第三步是怎么求的?再答:
log2(log3(log4x)=0(log3(log4x)=2^0=1log4x=3^1=3x=4^3=64log3(log4(log2y)=0log4(log2y)=3^0=1log2y=4^1=
=0+log2(3/12)=log21/4=-2
同底的对数相加,结果等于真数积的对数真数之积为cosπ/9cos2π/9cos4π/9=(8sinπ/9cosπ/9cos2π/9cos4π/9)/(8sinπ/9)=4sin2π/9cos2π/9c
log以4为底8的对数-log以9分之1为底3的对数-log以根号2为底4的对数=lg8/lg4-lg3/lg(1/9)-lg4/lg(√2)=3lg2/2lg2-lg3/(-2)lg3-2lg2/(
log2(log(0.5)(log(3)(81)))=log2(log(0.5)(4))=log2(-2)真数不能小于0,所以可能是题目出错了...请楼主检查
6手机难打对数,用换底功式,等我回家用电脑打给你过程?
log以3为底2的对数log以2为底3的对数>log以2为底2的对数=1>log以3为底2的对数
2log5(10)+log5(0.25)=2log5(2*5)+log5(1/4)=2[log5(2)+1]-log5(2^2)=2lo5(2)+2-2log5(2)=2
log2(6)-log2(3)=log2(6/3)=log2(2)=1
解;可以利用换底公式,变为相同的底数来计算原式=(log以2为底3为真数÷log以2为底4为真数)×(log以2为底2为真数÷log以2为底9为真数)×(log以2为底4√32为真数÷log以2为底1
log以2为底4+log以9为底的27-2log以2为底的3=lg4/lg2+lg27/lg9-2lg3/lg2=2lg2/lg2+3lg3/2lg3-2lg3/lg2=2+3/2-2lg3/lg2=
根据换底公式和对数运算法则,通式为:log(2^n)3^n=n/n*log₂3=log₂3∴[log₂3+log(4)9+log(8)27+……+log(2^n)3
log以2为底25乘以log以3为底2√2乘以log以5为底9=2log25×(3/2)log32×2log53=6log25×log32×(log23/log25)=6(log25×1/log52)
【参考答案】log5(3x-2)>log5(x+1)即不等式组:3x-2>0①x+1>0②3x-2>x+1③解这个不等式组得x>2/3x>-1x>3/2∴原不等式组的解集是x>3/2有不理解的地方欢迎
比较大小?
解题思路:本题柱考察学生对于对数的运算的理解和应用。解题过程:
用对数换底公式.log23*log98=(lg3/lg2)*(lg8/lg9)=(lg3/lg2)*(3lg2/2lg3)=3/2