新题库--第二章第08节对数与对数运算

1对数与对数运算题型1对数及对数运算1．把下列指数式写成对数式：(1)32＝８；（2）52＝32；（3）12＝21；（4）312731。解：(1)2log８＝３(2)2log32＝５(3)2log21＝－１(4)27log31＝－312．把下列对数式写成指数式：（1）416log21；（2）2log128=7；（3）lg0.01=-2；（4）ln10=2.303。解：（1）16)21(4（2）72=128；（3）210=0.01；（4）303.2e=103．用logax,logay,logaz,loga(x+y),loga(x-y)表示下列各式：（1）logazyx23；（2）loga(x423yz)；（3）loga(xy21z32)；（4）22ayxxylog；（5）y)yxyx(loga；（6）3a]y)x(xy[log．解：（1）logay2logxlog31z)logy(2logxlog31zylogxlogzyxaaaaa2a3a23zloga；（2）loga(x·423yz))ylogz(log41xlogyzlogxlog2a3aa423aazlog43ylog21xlogzlog43ylog42xlogaaaaaa；（3）loga(xy21z32)zlog32ylog21xlogzlogylogxlogaaa32a21aa；（4）y)y)(x(xlogylogxlog)y(xlogxylogyxxylogaaa22aa22ay)-(xlogy)(xlogylogxlogaaaa；（5）ylogy)(xlogy)(xlogylogyxyxlogy)yxyx(logaaaaaa；（6）y)(x3logx3logy3logy)](xlogxlogy3[log]y)x(xy[logaaaaaa3a。4．计算：（1）2log510+log50.25；（2）2log525+3log264；（3）log2(log216)解：（1）2log510+log50.25=log5102+log50.25=log5（100×0.25）=log525=2．2（2）2log525+3log264=2log552+3log226=2×2+3×6=22．（3）log2(log216)=log2(log224)=log24=log222=2．5．不查表求值：（1）42log2112log487log222．（2）)223223(log2．解：（1）原式3log217log21)327(log21)23(log237log212222242212log213log212log217log212log23log2log22222222．方法二：原式2121log4248127log22.（2）原式]1212[log])12()12([log22223)2(log)22log322．方法二：原式22322326(log21)223223(log2122238log212．6．不查表求值：(1)lg14-2lg37+lg7-lg18(2)2.1lg10lg38lg27lg（3）2(lg2)lg2lg50lg25；（4）3948(log2log2)(log3log3)新疆源头学子小屋特级教师王新敞@[e-mail]王新敞特级教师源头学子小屋新疆解：(1)lg14-2lg37+lg7-lg18=lg(2×7)-2(lg7-lg3)+lg7-lg(23×2)=lg2+lg7-2lg7+2lg3+lg7-2lg3-lg2=0奎屯王新敞新疆方法二：lg14-2lg37+lg7-lg18=lg14-lg2)37(+lg7-lg18=lg01lg18)37(7142奎屯王新敞新疆评述：此题体现了对数运算性质的灵活运用，运算性质的逆用常被学生所忽视.1023lg)10lg(32lg)3lg(2.1lg10lg38lg27lg)2(22132132312lg23lg)12lg23(lg23奎屯王新敞新疆（3）原式22(lg2)(1lg5)lg2lg5(lg2lg51)lg22lg5(11)lg22lg52(lg2lg5)2新疆源头学子小屋特级教师王新敞@[e-mail]王新敞特级教师源头学子小屋新疆（4）原式lg2lg2lg3lg3lg2lg2lg3lg3()()()()lg3lg9lg4lg8lg32lg32lg23lg23lg25lg352lg36lg24。7．求53log219的值.解：25353)3(3)3(3999225log5log25log215log213333.8．求值:log6[log4(log381)].3解：原式=log6[log4(log33)4]=log6(log44)=log61=0．9．用xalog，yalog，zalog表示下列各式：zxya(1)log32log)2(zyxa解：（1）zxyalog=alog（xy）-alogz=alogx+alogy-alogz.（2）32logzyxa=alog（2x3log)zya=alog2x+alog3logzya=2alogx+zyaalog31log21.10．计算：⑴27log9；⑵81log43；⑶32log32；⑷625log345；(5)2log（2log16）；（6）3log12.05；（7）4219432log2log3log。解：⑴设x27log9，则,279x3233x,∴23x；解法二：239log3log27log239399；⑵设x81log43，则8134x,4433x,∴16x；解法二：16)3(log81log1643344；⑶令x32log32=13232log,∴13232x,∴1x；解法二：32log32=132log132；⑷令x625log345,∴625534x,43455x,∴3x；解法二：3)5(log625log334553434；(5)2log（2log16）＝2log（2log42）＝2log４＝2log22＝２；（6）原式=15315555531log3log52.0；（7）原式=2345412log452log213log21232。11．计算下列各式：（1）log12.5-lg85+lg0.5（2）2lg3.0lg211000lg8lg27lg解：（1）原式21lg210lg2100lg43=lg100-lg23-lg10+lg24+lg1-lg2=lg102-3lg2-1+4lg2-lg2=2-1=1．（2）原式3)12lg23(lg21)12lg23(lg232lg)10lg3(lg2110lg232lg33lg23．412．化简27lg81lg3lg27lg539lg523lg解：原式491491lg3lg3lg3lg3(1)lg311510251024lg33lg3(43)lg35．方法二：原式5113lg3lg2781lg)32793lg(51121532152．题型2对数运算的综合问题13．（1）已知2log3=a，3log7=b,用a,b表示42log56；（2）已知18log9=a,b18=5,用a,b表示36log45。解：（1）∵2log3=a，则2log13a,又∵3log7=b,∴1312log7log2log37log42log56log56log33333342babab。（2）∵18log9=a，∴a2log1218log1818，∴18log2=1a，∵b18=5，∴18log5=b，∴aba22log15log9log36log45log45log181818181836。14．如果log8a+log4b2=5,log8b+logaa2=7，求log2(ab)的值。解：∵log8a+log4b2=5,log8b+logaa2=7,∴(logaa+log8b)+(log4b2+log4a2)=12,∴log8(ab)+log4(ab)2=12,∴log8(ab)+2log4(ab)=12,∴124log(ab)log28log(ab)log2222,∴31log2(ab)+log2(ab)=12,∴34log2(ab)=12,∴log2(ab)=12×43=9。15.已知3a=5b=A，且2b1a1，求A的值.解：∵3a=5b=A，∴a=log3A,b=log5A，∴5logb13,loga1AA，∵2b1a1，∴logA3+logA5=2,∴logA3×5=2,∴A2=15,∴A=±15，又A0,∴A=15.16．如果log8a+log4b2=5,log8b+logaa2=7，求log2(ab)的值.解：∵log8a+log4b2=5,log8b+logaa2=7,∴(logaa+log8b)+(log4b2+log4a2)=12,∴log8(ab)+log4(ab)2=12,∴log8(ab)+2log4(ab)=12,∴124log(ab)log28log(ab)log2222,5∴31log2(ab)+log2(ab)=12,∴34log2(ab)=12,∴log2(ab)=12×43=9.17.（1）已知log147=a,log145=b，求log3528．（2）已知log52=a，求2log510+log50.5的值；（3）已知log312=a，试用a表示log324．解：（1）log3528=log35(4×7)=log3522+log357=2log352+log357=2log35714+log357=2log3514-2log357+log357=2log3514-log357=baa27log5log7log275log7log235log7log35log14log2141414141414141414．（2）2log510+log50.5=2log5(5×2)+log521=2(log55+log52)+log52-1=2+2log52-log52=2+log52=2+a．（3）∵log312=log33×22=log33+log322=1+2log32，由log312=a得1+2log32=a,∴log32=21a，又log324=log3(3×23)=log33+log323=1+3log32=1+3213a21a．18.若)2(21lglgbagba，求ba4log之值。解：)2lg(2lglgbaba，2)2lg(lgbaab，054,)2(222abbabaab，4,1,295bababa或即.1,02,1,0,0babababa故则若舍去.1log,44baba.19．设x、y、z∈R+，且3x=4y=6z。（1）求证：yxz2111；（2）比较3x,4y,6z的大小。解：（1）设3x=4y6z=k(k0)，∵x