a=log18(9)=log2(9)/log2(18)=2log2(3)/[1+2log2(3)]
b=log18(5)=log2(5)/log2(18)=log2(5)/[1+2log2(3)]
由此解得 log2(3)=a/[2(1-a)],log2(5)=b/(1-a),
所以,根据对数换底公式得
log30(36)=log2(36)/log2(30)=[2+2log2(3)]/[log2(3)+1+log2(5)]=[2+a/(1-a)]/[a/(2(1-a))+b/(1-a)+1]=(4-4a+2a)/(a+2b+2-2a)=(4-2a)/(2b-a+2)