数学竞赛中的指数和对数函数问题

(,310036)f(x)=ax(a0,a1),R.f(x)=logax(a0,a1),R+,,y=x.,,,.11log627=a,log1816.:,.,2,log1816=log216log218=41+2log23(1)log627=log227log26=3log231+log23=a,a(1+log23)=3log23,log23=a3-a,(1)log1816=41+2a3-a=4(3-a)3+a.2a0,a1,loga3a+22a2+loga(3-8)a.loga3a+22a2=logaa3+22=logaa+loga3+22=12+loga(1+2),loga(3-8)a=logaa+loga3-22=12+loga(2-1),,loga3a+22a2+loga(3-8)a=12+loga(1+2)+12+loga(2-1)=1+loga(2+1)(2-1).=1+loga1=1+0=13a,b,c(abc)x,y,z,,ax=by=cz=70,1x+1y+1z=1.alogb(c-a).ax=by=cz=70,a1=701x,b1=701y,c1=701z,,(abc)1=701x+1y+1z.1x+1y+1z=1,abc=70.x,y,z,,ax=by=cz=701,a,b,c1,1abc.70=257,,a=2,b=5,c=7.alogb(c-a)=2log55=2.,.1(1998)a1,b1,lg(a+b)=lga+lgb,lg(a-1)+lg(b-1)()(A)lg2.(B)1.(C)0.(D)a,b.2(1986)f(x)=4x4x+2,f11001+f21001++f10001001.24a1,mp0,x+logax=mp,x+ax=m.:x+logax=mlogax=m-x(1)x+ax=max=m-x(2)141,(1)y=logaxy=m-x,(2)y=axy=m-x,y=logaxy=ax,,p(1)m-p(2).,,,.5f(x)=logb(x+x2-2)f-1(x)(b0,b1).1)f(x)f-1(x),f-1(x);2)P(n)=22f-1(n+logb2).P(n)3n+3-n2(nN+),b.:f-1(x)f(x).1)x+x2-20x2-20x2-2-x|x|2]x2.f(x)x2.,u(x)=x+x2-2(x2),u(x)u(2)=2.b1,f(x)[logb2,+);0b1,f(x)(-,logb2].b1,f-1(x)[logb2,+);0b1,f-1(x)(-,logb2].y=logb(x+x2-2),by=x+x2-2,by-x=x2-2.,(by)2-2xby+2=0.x=12(by+2b-y).f-1(x)=12(bx+2b-x)2)P(n)=22f-1(n+logb2)=24(bn+logb2+2b-n-logb2)=12(bn+b-n).P(n)3n+3-n212(bn+b-n)12(3n+3-n),(bn-3n)+(b-n-3-n)0,13nbn(3n-bn)[1-(3b)n]0,3n-bn01-(3b)n0(1)3n-bn01-(3b)n0(2)(1)13b3b1,(2).b(13,1)(1,3).3a,blog2x+x-3=02x+x-3=0,a+b.4f(x2-3)=logax26-x2(a0a1),f(x)f-1(x),f-1(x)0x0,a.3.6x,y,z,45x+9y+4z-6825x+9y+4z+256=0.x+y+z.:,u=25x+9y+4z,u2-68u+256=0,u1=4,u2=64.u=25x+9y+4z=4,5x+9y+4z=4.4(x+y+z)5x+9y+4z9(x+y+z),x+y+z1(x=y=0,z=1);x+y+z49(x=z=0,y=49).x+y+z49,1.u=25x+9y+4z=64,5x+9y+4z=36.,x+y+z4,9.,,x+y+z49,9,4.7(2001)1log12x+232.1log12x+232,1log12x+2321log12x+2-32,1log12x-121log12x-72.log12x-2log12x0-27log12x0,x40x11x227.(0,1)(1,227)(4,+).5(1995)[x]x,lg2x-[lgx]-2=0.6(1996){x|-1log1x10-12,xN+}.1.(C).2.500.3.3.4.0a1.5.3.6.290-1.(:2002-09-29)