log4(3a 4b)=log2 ab^1 2,则a b的最小值
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√13-3x=x-1x²+x-12=0x=3或-4(不在定义域内,舍去)x=3
2能约去
log4(5-x)=1/2log2(5-x)=log2(5-x)½=log2(x-2)因为log2x为单调函数,所以(5-x)½=(x-2)解得x=(3±√13)/2又因为x定义域
1、log2(4*5)-(2/2)log2(5)=log2(4)+log2(5)-log2(5)=2;2、换底公式:(lg3/lg2)*(lg5/lg2)*(lg5/lg3)=1;3、log2(5-l
应运换底公式㏒aN=(㏒bN)/(㏒ba)将上述式子转换(Log43Log83)*(Log32Log92)=[log43(log43)/(log48)]×[log32(log32/log39)]=(l
alog23<b=log32<c=log46再问:为什么c>b再答:哦,开始以为同底呢,应该这样:b=log32<1a=log23>1c=log46>1a=log23=lg3/lg2c=log46=l
log2^x=log4^9=log2^2(3^2)=log2^3所以x=3
如果log4(log2^x)是表示外面的log以4为底,而里面的log以10或其它数(>0)为底的话,则log4(log2^x)=[log2(log2^x)]/2=log2(log4^x)=log2(
log2(3)-log4(6)=log2(3)-log2^2((√6)^2)=log2(3)-log2(√6)=log2(3/√6)=log2(√3/√2)=log2(√3)-log2(√2)=1/2
log4(6)=1/2log2(6)=log2(√6)log4(6)
log43=log2^23=1/2log23=p/2
log2^3×log3^4×log4^x=log9^3(lg3/lg2)×(lg4/lg3)×(lgx/lg4)=log9^(9^1/2)所以lgx/lg2=1/2lgx=1/2*lg2=lg√2x=
log2(3-x)-log4(x+5)=1log2(3-x)=1+log4(x+5)log4(3-x)²=log44(x+5)(3-x)²=4(x+5)x²-6x+9=4
利用换底公式:log(a)(b)=log(n)(b)/log(n)(a):log2(3)*log3(4)*log4(5)*.log(k+1)(k+2)=log2(3)*[log2(4)/log2(3)
我的对,看清楚log2^x>=log4^(3x+4)==>log2^x>=log2^(3x+4)/log2^4==>2log2^x>=log2^(3x+4)==>log2^(x^2)>=log2^(3
3再问:怎么算的呢······重点是2log3(2)×log4(3),前面我会算再答:运用换底公式即可
因为log2^3=a所以log4^9=log(2^2)^3*3=log(2^3)^3*2=3*2log2^3=6a
由换底公式log4(x)=log2(x)/log2(4)=log2(x)/[2log2(2)]=(1/2)×log2(x)=log2(√x)∴log2(3X+2)-log2(√x)=3即log2[(3
=lg243/lg4-lg729/lg8=5lg3/2lg2-6lg3/3lg2=5lg3/2lg2-4lg3/2lg2=lg3/lg4=log4(3)原式=lg12/lg4-lg6/lg2=(2lg