logbx=3,logcx=6,则logabcx的值为______.答案1解析logabcx=1logxabc=1logxa+logxb+logxc,∵logax=2,logbx=3,logcx=6,∴logxa=12,logxb=13,logxc=16,∴logabcx=112+13+16=11=1.11.若26a=33b=62c,求证:1a+2b=3c.证明设26a=33b=62c=k(k>0),那么6a=log2k,3b=log3k,2c=log6k,∴1a=6log2k=6logk2,1b=3log3k=3logk3,1c=2log6k=2logk6.10∴1a+2b=6·logk2+2×3logk3=logk(26×36)=6logk6=3×2logk6=3c,即1a+2b=3c.12.设a>1,若对于任意的x∈[a,2a],都有y∈[a,a2]满足方程logax+logay=3,求a的取值范围.解∵logax+logay=3,∴logaxy=3,∴xy=a3,∴y=a3x.∵函数y=a3x(a>1)在[a,2a]上为减函数,又当x=a时,y=a2,当x=2a时,y=a32a=a22,∴a22,a2?[a,a2],∴a22≥a,又a>1,∴a≥2,∴a的取值范围为a≥2.三、探究与创新13.设x,y,z均为正数,且3x=4y=6z.(1)试求x,y,z之间的关系;(2)求使2x=py成立,且与p最接近的正整数(即求与p的差的绝对值最小的整数);(3)比较3x,4y,6z的大小.解(1)设3x=4y=6z=t,由x>0,知t>1,故取以t为底的对数,得xlogt3=ylogt4=zlogt6=1,∴x=1logt3,y=1logt4,z=1logt6,1z-1x=logt6-logt3=logt2=12logt4=12y,∴x,y,z之间的关系为1z-1x=12y.
全文预览
