a=log以3为底2的对数,bin2

log3(4)=lg4/lg3=2lg2/lg3=2a/blog2(12)=lg12/lg2=(lg3+2lg2)/lg2=(2a+b)/a

答：log3(7)=a,3^a=7log2(3)=b,2^b=3所以：(2^b)^a=7所以：2^(ab)=7所以：log2(7)=log2[2^(ab)]=ab所以：log2(7)=a

显然C>b>a

换底公式2a/b12=3*4对数基本性质b/a+2

a=lg3/lg2lg3=alg2b=lg7/lg3lg7=blg3=ablg2log42(56)=lg56/lg42=lg(2^3*7)/lg(2*3*7)=(3lg2+lg7)/(lg2+lg3+

lon1256=(log356)/(log312)=(log37*8)/(log33*4)=(log37+log38)/(log33+log34)=(b+log32^3)/(1+log32^2)=(b

a=lg3/lg2lg2=lg3/ab=lg5/lg3lg3=lg5/blg2=lg5/ablog以15为底20=lg20/lg15=(lg2+lg2+lg5)/(lg3+lg5)=(ab+2)/(a

答：a=log2(3),b=log3(7)=log2(7)/log2(3)所以：ab=log2(7)log14(56)=log2(7*8)/log2(14)=[log2(7)+3]/[log2(7)+

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）

底数3>1是增函数所以真数越大则对数值越大因为π>√3>√2所以log3(π)>log3(√3)>log3(√2)

由loga(b)=1/logb(a)又loga(x)=2,logb(x)=3,logc(x)=6,∴logx(a)=1/2,logx(b)=1/3,logx(c)=1/6logabc(x)=1/log

log以a为底x的对数=2∴log以x为底a的对数=1/2log以b为底x的对数=3∴log以x为底b的对数=1/3log以c为底x的对数=6∴log以x为底c的对数=1/6log以abc为底x的对数

log3(10)=a,log3(2)+log3(5)=a,log3(25)/log3(6)=b,[2log3(5)]/[log3(2)+1]=b,log3(2)=(a-b)/(b+1),log3(5)

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

log27(25)=(2/3)log3(5)=b所以log3(5)=3b/2用换底公式log10（5）=[log3(5)]/[log3(10)]=3b/2a

a=log1/3(2),b=log2(3),c=(1/2)^0.3a=log1/3(2)1,c=(1/2)^0.3c>a

log以2为底根号3=1/2(log以2为底3),log以3为底根号2=1/2(log以3为底2),log以3为底21故b>c

用换底公式log(A)M=log(b)M/log(b)A(b>0且b≠1）a=log(6)π/log(6)3>1b=log(6)6/log(6)7则0

a^2>b>a>1a/b1logb(a)=logb(a^2/a)>logb(b/a)>0logb(a)1loga(a/b)

log以a为底x的对数=2∴log以x为底a的对数=1/2log以b为底x的对数=3∴log以x为底b的对数=1/3log以c为底x的对数=6∴log以x为底c的对数=1/6log以abc为底x的对数