對(duì)數(shù)的運(yùn)算性質(zhì)+第1課時(shí)練習(xí) 高一上學(xué)期數(shù)學(xué)蘇教版（2019）必修第一冊(cè)

4.2.2對(duì)數(shù)的運(yùn)算性質(zhì)第1課時(shí)對(duì)數(shù)的運(yùn)算性質(zhì)一、選擇題1.log42-log48= ()A.-2 B.-1 C.1 D.22.若lga-2lg2=1,則a= ()A.4 B.10C.20 D.403.若10m=2,10n=6,則n-2m= ()A.-lg2 B.lg2 C.-lg3 D.lg34.log(n+1-n)(n+1+A.1 B.-1 C.2 D.-25.已知lga,lgb是方程6x2-4x-3=0的兩根,則lgba2=A.49 B.139 C.149 6.如果lgx=lga+3lgb-5lgc,那么 ()A.x=a+3b-5c B.x=3C.x=ab3c5 D7.若lg2=m,lg3=n,則lg12lg15= (A.2m+n1+m+C.2m+n1-m8.(多選題)以下運(yùn)算錯(cuò)誤的是 ()A.lg2×lg3=lg6 B.(lg2)2=lg4C.lg2+lg3=lg5 D.lg4-lg2=lg29.(多選題)若ab>0,則下列各式中,一定成立的是 ()A.lg(ab)=lga+lgb B.lgab=lgalgC.12lgab2=D.lg6(ab)2=二、填空題10.方程lgx+lg1x+9=1的解為11.已知a=log32,那么log38-2log36的計(jì)算結(jié)果可以用a表示為.

12.若方程(lgx)2+(lg5+lg7)lgx+lg5·lg7=0的兩根分別是α,β,則αβ的值是.

三、解答題13.用lnx,lny,lnz表示下列各式:(1)lnxyz;(2)ln(x2y4z3);(3)ln(xy×3z14.計(jì)算下列各式的值:(1)lg5+log36+lg20-log32;(2)(lg5)2+lg2·(1+lg5)-eln2.(3)(lg15.若lg(x-y)+lg(x+2y)=lg2+lgx+lgy,則xy=16.已知lg2=a,lg3=b.(1)求lg72,lg4.5(用a,b表示);(2)若lgx=a+b-2,求x的值.4.2.2對(duì)數(shù)的運(yùn)算性質(zhì)第1課時(shí)對(duì)數(shù)的運(yùn)算性質(zhì)1.B[解析]log42-log48=log414=-1,故選B2.D[解析]將lga-2lg2=1化為lga-lg22=1,即lga4=1,所以a4=10,解得a=3.D[解析]∵10m=2,10n=6,∴m=lg2,n=lg6,∴n-2m=lg6-2lg2=lg6-lg2=lg62=lg3,故選D4.B[解析]log(n+1-n)(n+1+n)=-log(n+15.D[解析]∵lga,lgb是方程6x2-4x-3=0的兩根,∴l(xiāng)ga+lgb=23,lga·lgb=-12,∴l(xiāng)gba2=(lga-lgb)2=(lga+lgb)2-4lga·lgb=2326.C[解析]lgx=lga+lgb3-lgc5=lgab3c57.C[解析]lg12lg15=lg(3×228.ABC[解析]lg2+lg3=lg6,故A,C中運(yùn)算錯(cuò)誤;lg2+lg2=lg4,則lg4-lg2=lg2,故B中運(yùn)算錯(cuò)誤,D中運(yùn)算正確.故選ABC.9.CD[解析]對(duì)于A,當(dāng)a<0,b<0時(shí),等式右邊無意義,A不一定成立;對(duì)于B,當(dāng)a<0,b<0時(shí),等式右邊無意義,B不一定成立;對(duì)于C,∵ab>0,∴12lgab2=lgab,C一定成立;對(duì)于D,∵ab>0,∴l(xiāng)g6(ab)2=lg(ab)10.x=1[解析]∵lgx+lg1x+9=lg(1+9x)=1,∴1+9x=10,解得x=11.a-2[解析]log38-2log36=3log32-2×(log32+1)=a-2.12.135[解析]由已知得lgα+lgβ=-(lg5+lg7),則lg(αβ)=lg135,解得αβ=13.解:(1)lnxyz=lnx+lny-ln(2)ln(x2y4z3)=lnx2+lny4+lnz3=2lnx+4lny+3lnz.(3)ln(xy×3z)=lnxy+ln3z=12lnx+12ln14.解:(1)原式=(lg5+lg20)+(log36-log32)=lg100+log33=2+1=3.(2)(lg5)2+lg2·(1+lg5)-eln2=(lg5)2+lg2+lg2·lg5-2=lg5·(lg5+lg2)+lg2-2=lg5+lg2-2=-1.(3)原式=(lg3(1-lg315.2[解析]根據(jù)對(duì)數(shù)的運(yùn)算法則知,lg[(x-y)(x+2y)]=lg(2xy),則x2-2y2=xy,即xy2-xy-2=0,又由題意知x>0,y>0,∴x16.解:(1)lg72=lg(2