log8底3=p,log3底5=9,用p.q表示lg5

备注：第一个（）内是“底数”,第二个（）内是“真数”log8底9*log3底32=log（3的平方）（2的三次方）*log（2的五次方）（3)=3/2log(3)(2)*1/5log(2)(3)=3/

log(8)3=lg3/3lg2=a,log(3)5=lg5/lg3=b,lg2+lg5=1lg5=3ab/(1+3ab)

log(8)9=lg9/lg8=2lg3/3lg2=a,lg2=2lg3/3a.log(3)5=lg5/lg3=b,lg3=lg5/b.lg5+lg2=lg10=1所以,lg2=2/(3ab+2)

=(log(2,3)/2+log(2,3)/3)*(log(3,2)+1/3-?)log(a,b)=ln(b)/ln(a)

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)

=2log2(5)×2log3(2)×2log5(3)=8(lg5/lg2)(lg2/lg3)(lg3/lg5)=8=[1/2log2(3)+1/3log2(3)][log3(2)+1/2log3(2

a=(lg3)/(lg8)b=(lg5)/(lg3)ab=lg8/lg5=(3lg2)/(lg5)=3(lg(10/5)/(lg5))=3(lg10-lg5)/lg5=3(1-lg5)/lg5所以lg

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)

(log23/log24+log23/log28)(log35/log33+log35/log39)(log52/log55+log52/log525)=[(5log23)/6][(3log35)/2

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

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

a=log89=lg9/lg8=2lg3/3lg2b=log320=lg20/lg3=(1+lg2)/lg3ab=2(1+lg2)/3lg2=(2/3lg2)+1/3ab-1/3=2/3lg2lg2=

m=log3(5)=lg5/lg3n=log8(3)=lg3/lg8=(lg3/lg2)/3lg5=(lg5/lg3)*(lg3/lg2)*lg2=3mn/lg2lg5lg2=lg5(1-lg5)=3

n=log(8)3=log(2^3)3=1/3log(2)3=1/3lg3/lg2(换底公式）m=log(3)5=lg5/lg3上面两式相乘有pq=1/3(lg5/lg2)3pq=lg5/lg2=lg

log83=a==>8^a=3log35=b==>3^b=58^(ab)=52^(3ab)=5∴log25=3ab2^(3ab+1)=10所以log210=3ab+1∵lg5=log105=(log2

log3(2)=lg2/lg3log5(3)=lg3/lg5log8(5)=lg5/lg8log3(2)*log5(3)*log8(5)=(lg2*lg3*lg5)/(lg3*lg5*lg8)=lg2

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=