对数的运算一、选择题1.已知ln2=a,ln3=b,那么log32用含a,b的代数式表示为()A.a-bB.C.abD.a+b2.若lgx-lgy=t,则lg3-lg3=()A.3tB.tC.tD.3.若2.5x=1000,0.25y=1000,则-=()A.B.3C.-D.-34.已知x,y,z都是大于1的正数,m>0,且logxm=24,logym=40,logxyzm=12,则logzm的值为()A.B.60C.D.5.已知a=log32,则log38-2log36的值是()A.a-2B.5a-2C.3a-(1+a)2D.3a-a2-1二、填空题6.=________.7.设x=log23,则=________.8.已知log23=a,log37=b,则log1456=________.三、解答题9.已知2x=3y=6z≠1,求证:+=.10.已知loga(x2+4)+loga(y2+1)=loga5+loga(2xy-1)(a>0,且a≠1),求log8的值.1答案课时跟踪检测(十七)1.选Blog32==.2.选Alg3-lg3=3lg-3lg=3lg=3(lgx-lgy)=3t.3.选A∵x=log2.51000,y=log0.251000,∴===log10002.5,同理=log10000.25,∴-=log10002.5-log10000.25=log100010==.4.选B由已知得logm(xyz)=logmx+logmy+logmz=,而logmx=,logmy=,故logmz=-logmx-logmy=--=,即logzm=60.5.选Alog38-2log36=3log32-2(log32+log33)=3a-2(a+1)=a-2.6.解析:=====1.答案:17.解析:法一:由x=log23得2x=3,2-x=,==32+3×+2=.法二:==22x+1+2-2x=32+1+=.答案:8.解析:由log23=a,log37=b,得log27=ab,则log1456====.答案:9.证明:设2x=3y=6z=k(k≠1),∴x=log2k,y=log3k,z=log6k,∴=logk2,=logk3,=logk6=logk2+logk3,∴=+.10.解:由对数的运算法则,可将等式化为loga[(x2+4)·(y2+1)]=loga[5(2xy-1)],∴(x2+4)(y2+1)=5(2xy-1).整理,得x2y2+x2+4y2-10xy+9=0,配方,得(xy-3)2+(x-2y)2=0,∴∴=.∴log8=log8=log2-1=-log22=-.2