2、-4a9-4■A解析:原式=1g31g3ilg2Ig2、彳+igi母+的Oa1,JgA.'lg^,JgA.2lg2十3lg21g3十2lg3111g3=12+3卜1glx,i/ig251+2Jigl=4.答案：B4l富的值是()A.2B.3C.1D.223解析:21og491ogz721-7=103-H1og27=".1ogz71ogz223答案：D5.若1g2=a,1g3=b,则露2等于(2a+bA1—a+b2a+bB1+a+ba2bC.d1—a+ba+2bD.1+a+b解析:1g121g3+1g41g3+21g22a+b1g15=1g3+1g5=1g3+(1—1g2)=1-
3、a+b.答案：A二、填空题1…6.已知m>0,且10=1g(10m)+1gm，则x=1解析：因为1g(10m)+1gm=111一"1g10mm!=1g10=1,所以10x=1,得x=0.答案：01一一一7.右f(x)=ax—Q且f(1ga)={10,则a=解析：f(1ga)=a1ga-2=aja=V10>1所以a1ga=(10a)2,两边取对数,1彳#(lga)=2(1+lga),1所以2(lga)2—lga—1=0,解得1ga=1或1ga=—Q,则a=10或a=V0.答案：10或彗21g4+lg98.11—1+2©0.36+31g8解析：原式=2(lg4+lg3)1+lg/0
4、36+1g3821g121+lg0.6+lg221g12lg(10X0.6X2)=2.答案：2三、解答题9.计算：(1)71-1og75;1(2)100(21g9-lg2);(3)alogabTogbc(a,b为不等于1的正数,c>0).解：(i)原式=7><7—log75=7og5=5.(2)原式=1002lg9x1001g2=101g9x100M八19=9(101g2)2=4.(3)原式=(alogab)logbc=blogbc=c.7.已知1gx+1gy=2lg(x—2y),求log&j的值解：由lgx+lgy=21g(x—2y),得1g(xy)=1g(x-2y)2,2x
5、xy=(x—2y),x>0,从而有y>0,x-2y>0,由①得x2—5xy+4y2=0,即(x—y)(x—4y)=0.x由②③④知x-y>0,故x—4y=0,即y=4.所以logj2xy=log,24=log2(72)4=4.B级能力提升1.计算log225logs2&log59的结果为（）A.3B.4C.5D.63解析:lg251g22lg921g521g221g3八原式一lg21g31g5=1g21g31g5=$答案:2.已知logu7=a,logu5=b,则用a,b表示log35l4=解析：*14=般==.logi47+logi45a+b一1答案：a^b3.已知log8
6、9=a,18b=5,用a,b表示log3645的值.log189+log1851+log182解析：方法一：因为10g189=a,18b=5,所以10g185=b.干曰1c.4匚log1845log18(9X5)于无log3645-log1836-log18(18X2)a+ba+b于是1。项645=logi8(9X5)10g8*1：210989+10985a+b210g1818—10g1892—a方法三：因为10g189=a,18b=5,所以lg9=alg18,lg5=blg18.所以log3645=lg45lg(9X5)lg9+lg51g36=9一2=——1822lg18-l
7、g99alg18+blg18a+b=2lg18-alg18=2a.18=2—a.1+log189方法二：因为10g189=a,18b=5,所以10g185=b.
