若log8∧3等于p

=(9*log8)/(3*log2)=(9*log2^3)/(3*log2)=(9*3*log2)/(3*log2)=9

log4(3)+log8(3)=（1/2）log2(3)+（1/3）log2(3)=（5/6）log2(3)

log8(9)=1og2^3(3^2)=2/3*log2(3)∴log8(9)/log2(3)=2/3*1=2/3

运用换底公式：log827除以log23=(ln27/ln8)/(ln3/ln2)=(3ln3/3ln2)*(ln2/ln3)=(ln3/ln2)*(ln2/ln3)=1再问：嗯我这块有点小不会有些题

log4(16)=2log3(4)*log4(8)*log8(m)=log3(m)=2m=9再问：为什么log3(4)*log4(8)*log8(m)=log3(m)？？？？？？？？？？？？？？？/呢

log8^3=p,→lg3/3lg2=p→lg3=p*3lg2log3^5=q→lg3/lg5=q→lg3=q(1-lg2)∴p*3lg2=q(1-lg2)lg2=q/(q-3p)

方法一(log89/log23)=(lg9/lg8)/(lg3/lg2)=(2lg3/3lg2)/(lg3/lg2)=(2/3)*(lg3/lg2)/(lg3/lg2)=2/3方法二(log89/lo

log83=p,lg3/lg8=p.log35=q,lg5/lg3=q.两式相乘得:lg5/lg8=pq.因lg8=lg2^3=3lg2=3×（1-lg5）所以lg5/[3×（1-lg5）]=pq.l

即p=lg3/lg8q=lg5/lg3所以pq=lg5/lg8=lg5/3lg2=lg5/[3(1-lg5)]所以lg5=3pq+pqlg5lg5=-3pq/(pq-1)

六分之五再问：过程，谢啦再答：log4(3)=1/2log2(3)log8(3)=1/3log2(3)故原式=5/6log2(3)Xlog3(2)=5/6

以10为底换地公式.log3/log8=p,log5/log3=qpq=log5/log8log5=pqlog8=3pqlog2

log4(16)=2log3(4)*log4(8)*log8(m)=log3(m)=2m=9再问：Thatyou,goodluck再问：lg8+3lg5=____答案不重要，过程，谢谢

上面的计算似乎不对可以这样算p=log(8)3=log(2^3)3=1/3log(2)3=1/3lg3/lg2(换底公式）q=log(3)5=lg5/lg3上面两式相乘有pq=1/3(lg5/lg2)

再答：已通知提问者对您的回答进行评价，请稍等

换底lg9/lg8=alg5/lg3=b相乘得ab=2lg5/3lg2则2lg5=3lg2xab1=lg10=lg2x5=lg2+lg52(1-lg2)=3lg2Xablg2=2/(2+3ab)

log4(3)+log8(3)=log2(3)/log2(4)+log2(3)/log2(8)=0.5log2(3)+1/3*log2(3)=5/6*log2(3)在高中阶段运算到这一步化为单个对数就

这个问题是这样的4=2²=（2^3）^(2/3)=8^(2/3)所以log8(8^(2/3))=2/3前面的是对数运算法则logaM-logaN=loga(M/N)

log2^3+log8^9=log2^3+(2/3)log2^3=(5/3)log2^3

log8(9)*log2(3)=lg9/lg8*lg3/lg2=2/3lg3的平方/lg2的平方

哦.由已知可得：log8(3)=lg3/lg8=lg3/(3lg2)=a,log3(5)=lg5/lg3=b所以：[lg3/(3lg2)]×(lg5/lg3)=ab即lg5/(3lg2)=ablg5=