2024-2025學(xué)年新教材高中數(shù)學(xué)第4章對(duì)數(shù)運(yùn)算與對(duì)數(shù)函數(shù)2.1對(duì)數(shù)的運(yùn)算性質(zhì)鞏固練習(xí)含解析北師大版必修第一冊(cè)

PAGE2.1對(duì)數(shù)的運(yùn)算性質(zhì)課后訓(xùn)練·鞏固提升一、A組1.log242+log243+log244等于()A.1 B.2 C.24 D.1解析:log242+log243+log244=log24(2×3×4)=log2424=1.故選A.答案:A2.化簡(jiǎn)12log612-2log62的結(jié)果為(A.62 B.122 C.log63 D.1解析:原式=log612-log62=log6122=log63.故選C答案:C3.方程(lgx)2+(lg2+lg3)lgx+lg2lg3=0的兩根的積x1x2等于()A.lg2+lg3 B.lg2lg3 C.16 D.-解析:∵lgx1+lgx2=-(lg2+lg3),∴l(xiāng)g(x1x2)=-lg6=lg6-1=lg16∴x1x2=16.故選C答案:C4.21+12loA.2+5 B.25 C.2+52 D.1+解析:21+12log25=2×212log答案:B5.已知a=log32,那么log38-2log36用a表示為()A.a-2 B.5a-2C.3a-(1+a)2 D.3a-a2-1解析:由log38-2log36=3log32-2(log32+log33)=3a-2(a+1)=a-2.答案:A6.已知a23=49(a>0),則lo解析:∵a23=49,∴a2=64∴l(xiāng)og23a=log2答案:37.計(jì)算lg14-lg25÷100解析:lg14-lg25÷100-12=lg1100答案:-208.lg0.01+log216的值是.

解析:lg0.01+log216=lg1100+log224=-2+4=2答案:29.解方程(lgx)2+lgx5-6=0.解:原方程可化為(lgx)2+5lgx-6=0,即(lgx+6)(lgx-1)=0,所以lgx=-6或lgx=1,解得x=10-6或x=10.經(jīng)檢驗(yàn)x=10-6和x=10都是原方程的解,所以原方程的解為x=10-6或x=10.二、B組1.計(jì)算log34273+lg25+lg4+7logA.-14 B.4 C.-154 D解析:原式=log3427-log33+lg52+lg22+2=14log333-1+2lg5+2lg2+2=34-1+2+2答案:D2.已知函數(shù)f(x)滿意:當(dāng)x≥4時(shí),f(x)=12x;當(dāng)x<4時(shí),f(x)=f(x+1),則f(2+log23)=(A.124 B.112 C.18解析:因?yàn)?+log23<2+log24=4,3+log23>3+log22=4,故f(2+log23)=f(2+log23+1)=f(3+log23)=12答案:A3.若lga,lgb是方程2x2-4x+1=0的兩個(gè)實(shí)根,則lgab2的值為A.2 B.12 C.4 D.解析:lgab2=(lga-lgb)2=(lga+lgb)2-4lgalgb=22-4×1答案:A4.若lg2=a,lg3=b,則用a,b表示lg45=.

解析:lg45=12lg45=12lg(5×9)=12lg5+12lg9=12(1-lg2)+lg3=-12lg2+答案:-12a+b+5.已知2x=9,log283=y,則x+2y的值為.解析:由2x=9,得log29=x,所以x+2y=log29+2log283=log29+log2649=log264=答案:66.求下列各式的值:(1)log535+2log52-log515-log5(2)[(1-log63)2+log62·log618]÷log64;(3)lg5(lg8+lg1000)+(lg23)2+lg0.06+lg1解:(1)原式=log535+log52-log515-log514=log535×215×14=log535014=log(2)原式=log663=[(log62)2+log62(log636-log62)]÷log64=[(log62)2+2log62-(log62)2]÷log64=2log62÷log64=log64÷log64=1.(3)原式=lg5(3lg2+3)+3(lg2)2+lg6100-=lg5(3lg2+3)+3(lg2)2+lg6-2-lg6=3·lg5·lg2+3lg5+3·(lg2)2-2=3lg2(lg2+lg5)+3lg5-2=3lg2+3lg5-2=3(lg2+lg5)-2=3-2=1.7.已知f(x)=x2+(lga+2)x+lgb,f(-1)=-2,方程f(x)=2x至多有一個(gè)實(shí)根,求實(shí)數(shù)a,b的值.解:由f(-1)=-2得,1-(lga+2)+lgb=-2,所以lgba=-1=lg1所以ba=110,即又因?yàn)榉匠蘤(x)=2x至多有一個(gè)實(shí)根,即方程x2+(lga)x+lgb=0至多有一個(gè)實(shí)根,所以(lga)2-4lgb≤0,即[lg(10b)]2-4lgb≤0,所以(1-lgb)2≤0,所以lgb=1,b=10,從而a=100.故實(shí)數(shù)a,b的值分別為100,10.8.設(shè)a>1,若對(duì)于隨意的x∈[a,2a],都有y∈[a,a2]滿意方程logax+logay=3,求a的取值范圍.解:∵logax+l