已知log8底9=a

log8(a)+log4(b^2)=5===>1/3*log2(a)+log2(b)=5---------1)log8(b)+log4(a^2)=7===>1/3*log2(b)+log2(a)=7-

log89=a,log925=b,求lg22/3log2(3)=a,log3(5)=b,log2(3)=3a/2,log2(5)/log2(3)=b,log2(5)=b*log2(3)=3ab/2.l

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)

log89=lg9/lg8=2lg3/3lg2=a;log320=lg20/lg3=(1+lg2)/lg3=b;从此建立二元一次方程组,解得lg2=2/(3ab-2),

a=lg9/lg8=2lg3/3lg2lg2=2lg3/(3a)b=lg5/lg2=(1-lg2)/lg2=1/lg2-1b+1=1/lg2=3a/(2lg3)2lg3=3a/(b+1)lg3=3a/

由题意可知：a>0且b>0因为½log8（a）+log4（b）=5/2,所以：log8（a）+2log4（b）=5（1）又log8(b)+log4(a^2)=7,则：log8(b)+2log

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

a=log4(3)=lg3/lg4所以lg3=alg4=2alg2log8(9)=lg9/lg8=2lg3/3lg2=4alg2/3lg2=4a/3log4(3)=lg3/2lg2=(1/2)*log

sina=-2/3,因为a∈（-π/2,0）,所以cosa=√5/3cos(π+a)=-cosa=-√5/3

如果8,4是底数第一个方程等价于1/6log2（a）+1/2log2(b)=2.5(1)第二个等价于log2(a)+1/3log2(b)=7(2)(1)*2-(2)*31/3log2(a)-3log2

log89=2/3log23=mlog23=1.5mlog23*log35=log25=3/2*mnlog512=log5(2*2*3)=2*log52+log53=2*2/(3mn)+1/n=(4+

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

见图

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

log8^3=a/3,log8^9=2a/3

先介绍换底公式：log(a)b=lgb/lga证明：设log(a)b=t则a^t=b,两边取以10为底的对数lga^t=lgbtlga=lgb所以t=lgb/lga所以log(a)b=lgb/lga＝

（1）（2）（3）怎么用这个范围太大了,只能说因题而异再问：lg18等于log18吗再答：有的计算器默认log是lg再问：知道了谢谢再答：不客气

sin(π-a)=log8(1/4)={log2(1/4)}/{log2(8)=(-2)/3=-2/3sina=-2/3a属于（-π/2,0）tan(3π/2+a)=tan{2π-π/2+a)=tan