若log1 2 x=m,log1 4 y=m 2

运用换底log14(56)=log3(56)/log3(14)=log3(7*8)/log3(2*7)=〔log3(7)＋log3(8)〕/〔log3(7)＋log3(2)〕log3(2)=1/log

∵f（x）为（0，+∞）的单调函数，f(f(x)+log12x)＝6令f(x)+log12x=t，∴t为定值（单调）∴f（x）=log2x+t且f（t）=6∴log2t+t=6，∴log2t=6-t∴

log23=a;log37=b即lg3/lg2=a;lg7/lg3=b;所以lg7/lg2=a*b所以log72=1/a*b同理log142=1/1+ab;所以结果为（3+ab）/(1+ab)

log23=lg3/lg2=a,所以lg3=alg2,同理：log37=lg7/lg3=b,所以lg7=blg3即lg7=ab*lg2log1456=lg56/lg14其中lg56=lg7+lg8=l

运用换底:log14(56)=log3(56)/log3(14)=〔log3(7)＋log3(8)〕/〔log3(7)＋log3(2)〕log3(2)=1/log2(3)=1/alog3(8)=3lo

答：ab=log2^7ab+1=log2^7+1=log2^14ab+3=log2^7+3=log2^56log14^56=(ab+3)/(ab+1)

log14^56=log3^56/log3^14=(log3^7+3log3^2)/(log3^7+log3^2)=(b+3/a)/(b+1/a)=(3+ab)/(1+a)

若log23=a,log37=b则log23*log37=ab即log27=ablog1456=log214/(log256)=(log22*7)/(log27*8)=(log22+log27)/(l

∵函数y=|log0.5x|的值域为[0，2]，那么0≤log 12x≤2或-2≤log 12x≤0，∴14≤x≤1或1≤x≤4∴函数y=|log12x|的定义域区间长度b-a的最

log14(2)=alog2(14)=1/a=log2(2)+log2(7)log2(7)=1/a-1log√2(7)=2(1/a-1)

运用换底:log14(56)=log3(56)/log3(14)=〔log3(7)＋log3(8)〕/〔log3(7)＋log3(2)〕log3(2)=1/log2(3)=1/alog3(8)=3lo

(1)由换底公式得到log14^7=ln7/ln14=ln7/(ln2+ln7)=a;故ln7=aln2+aln7;解得ln7=aln2/(1-a);14^b=5.故log14^5=b.即ln5/(l

log14^2=lg2/lg14=lg2/(lg2+lg7)log16^49=lg49/lg16=2lg7/(4lg2)

a=lg3/lg2lg2=lg3/ab=lg7/lg3lg7=blg3log1456=lg56/lg14=(lg7+lg8)/(lg7+lg2)=(lg7+3lg2)/(lg7+lg2)=(blg3+

由换底公式：log14(2)=lg2/lg14log√2(7)=lg7/lg√2lg√2=(lg2)/2所以log√2(7)=lg7/lg√2=2lg7/lg2而：log14(2)=lg2/lg14=

第二题,做出来了!1.log147=lg7/lg14log145=lg5/lg14A=lg7/lg14=lg7/(lg7+lg2)B=lg5/lg14=(1-lg2)/(lg7+lg2)通过以上两式可

log35^2=log35^14-log35^7=1/(a+b)-1/log7^35=1/(a+b)-1/(1+b/a)=(1-a)/(a+b)原式=(1+1-a)/(a+b)=(2-a)/(a+b)

用换底公式求即可.log35（28）=log14(28)/log14(35)=[log14(14)+log14(2)]/[log14(7)+log14(5)]=[1+1-log14(7)]/[log1

要使函数有意义，则2+log12x≥0tanx≥0⇒0＜x≤4kπ≤x＜kπ+π2，得0＜x＜π2，或π≤x≤4，即函数的定义域为（0，π2）∪[π，4]．

log23=a----->log32=1/alog37=blog3(56)=log3(7*2^3)=log3(7)+3log3(2)=b+3/a=(ab+3)/alog3(14)=log3(7*2)=