作业帮已知log以8为底9的对数=m,log以3为底5的对数
来源：学生作业帮助网 编辑：作业帮 时间：2024/05/14 15:13:53
即lg4/lg3*lg8/lg4*lgm/lg8=lg16/lg4lgm/lg3=2lg4/lg4=2lgm=2lg3=lg3²所以m=9
a=lg27/lg12=3lg3/(lg3+2lg2)lg3+2alg2=3lg3lg3=alg2所以log6(16)=lg16/lg6=4lg2/(lg2+lg3)=4lg2/(lg2+alg2)=
[log9(4)+log3(8)]/log1/3(16)=[lg4/lg9+lg8/lg3]/[lg16/lg1/3]=[2lg2/2lg3+3lg2/lg3]/[4lg2/(-lg3)]=4lg2/
解由对数函数f(x)=(m²-4m+4)logm(x)知m^2-4m+4=1且m＞1且m＞0即m^2-4m+3=0且m＞1且m＞0即(m-3)(m-1)=0且m＞1且m＞0解得m=3故f(x
log15(5)=mlog15(3)=log15(15/5)=log15(15)-log15(5)=1-m
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/(
（M-1)分之（M-2）求采纳
=(lg3/lg4+;g3/lg8)(lg2/lg3+lg2/lg9)=(lg3/2lg2+;g3/3lg2)(lg2/lg3+lg2/2lg3)=(1/2+1/3)*lg3/lg2*(1+1/2)*
8∧a=3de2∧3a=3将3带入log35得log25=3abze则lg5/lg2=3ab且lg2+lg5=1可解得lg5=1/(3ab+1）
8再问：是不是换成分数形式可以互相约掉再答：log2(25)*log3(4)*log5(9)=lg25*lg4*lg9/lg2*lg3*lg4=log2(4)*log3(9)*log5(25)=2*2
即loga(m-2n)²=loga(mn)所以(m-2n)²=mn(m-n)(m-4n)=0m=n或m=4n真数大于0所以m-2n>0,m>0,n>0则m=n时不成立所以m=4nn
楼上写错了[log(3,2)+log(9,2)]*[log(4,3)+log(8.3)]=[log(3,2)+1/2log(3,2)]*[1/2log(2,3)+1/3log(2,3)]=3/2log
换底公式原式=(lg3/lg4+lg3/lg8)(lg2/lg3+lg2/lg9)=(lg3/2lg2+lg3/3lg2)(lg2/lg3+lg2/2lg3)=(lg3/lg2)(1/2+1/3)*(
log以3为底的4的对数/log以9为底的8的对数=log以3为底的2^2的对数/log以3^2为底的2^3的对数=log3(2^2)/log3^2(2^3)=4/3再问：最后一步在详细点再答：log
log89=a∴2lg3=3alg2lg3=3alg2/2log35=b∴lg5=blg31-lg2=blg3=3ablg2/22-2lg2=3ablg2∴lg2=2/(3ab+2)
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：
2log3(2)-log3(32/9)+log3(8)-5*2*log5(3)=log3(4)-log3(32/9)+log3(8)-5*2*log5(3)=log3(4/(32/9))+log3(8