a=log以5为底的3的对数 b=log以5为底的4的对数-搜答案-神马作文网

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

已知log18(9)=a;log18(5)=b;则有1=log18(18)=log18(2*9)=log18(9)+log(2)=a+log18(2);log18(45)=log18(9*5)=log

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

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)

1.假设其分别为a1,a2,a3先把a3化为同底数：a3=log以2为底1/3的对数设函数y=log以2为底x的对数,函数是单调递增的由1/3

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

(lg3/lg5)[(3lg5)/(3lg3)]=1

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

log以8为底9的对数=2/3log以2为底3的对数然后用换底公式,有关于log的一堆公式

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

log89=a∴2lg3=3alg2lg3=3alg2/2log35=b∴lg5=blg31-lg2=blg3=3ablg2/22-2lg2=3ablg2∴lg2=2/(3ab+2)

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