2024-2025學(xué)年新教材高中數(shù)學(xué)第四章指數(shù)函數(shù)對(duì)數(shù)函數(shù)與冪函數(shù)4.2.1對(duì)數(shù)運(yùn)算課時(shí)素養(yǎng)評(píng)價(jià)含解析新人教B版必修第二冊(cè)

PAGE對(duì)數(shù)運(yùn)算(15分鐘30分)1.在M=log3(x2-x-6)中,要使式子有意義,x的取值范圍是 ()A.x>3 B.x<-2C.x<-2或x>3 D.x<-3或x>-2【解析】選C.由題意,x2-x-6>0,解得x<-2或x>3.2.若x=QUOTE16,則x= ()A.-4 B.-3 C.3 D.4【解析】選A.x=QUOTE16=QUOTE=-4.3.若x=log43,則4x+4-x的值為 ()A.3 B.4 C.QUOTE D.QUOTE【解析】選D.因?yàn)樵?QUOTE+QUOTE=3+QUOTE=QUOTE.4.計(jì)算QUOTE8+log243=________.

【解析】原式=QUOTE+log226=-3+6=3.答案:35.若logπ[log2(lnx)]=1,則x=________.

【解析】由logπ[log2(lnx)]=1,所以log2(lnx)=π,所以lnx=2π,所以x=QUOTE.答案:QUOTE6.計(jì)算lg0.001+log282+QUOTE+lne-3.【解析】原式=lg10-3+log226+4×QUOTE-3=-3+6+QUOTE-3=QUOTE.(30分鐘60分)一、單選題(每小題5分,共20分)1.QUOTE-2QUOTE-lg0.01+lne3等于 ()A.14 B.0 C.1 D.6【解析】選B.原式=4-(33QUOTE-(-2)+3=4-9-(-2)+3=0.2.方程QUOTE=QUOTE的解是 ()A.x=QUOTE B.x=QUOTEC.x=QUOTE D.x=9【解析】選A.因?yàn)镼UOTE=2-2,所以log3x=-2,所以x=3-2=QUOTE.3.已知fQUOTE=QUOTE,則f(4)= ()A.QUOTElog25 B.QUOTElog23 C.QUOTE D.QUOTE【解析】選B.令2x+1=4,得x=log23,所以f(4)=QUOTElog23.4.設(shè)0<a<1,實(shí)數(shù)x,y滿意x+logay=0,則y關(guān)于x的函數(shù)的圖像大致形態(tài)是()【解析】選A.因?yàn)閤+logay=0,所以logay=-x,所以y=a-x,即y=(a-1)x=QUOTE,又因?yàn)?<a<1,所以QUOTE>1,所以指數(shù)函數(shù)y=QUOTE的圖像單調(diào)遞增,過(guò)點(diǎn)(0,1).二、多選題(每小題5分,共10分,全部選對(duì)的得5分,選對(duì)但不全的得3分,有選錯(cuò)的得0分)5.對(duì)于a>0,且a≠1,下列說(shuō)法中不正確的是()①若M=N,則logaM=logaN;②若logaM=logaN,則M=N;③若logaM2=logaN2,則M=N;④若M=N,則logaM2=logaN2.A.① B.② C.③ D.④【解析】選ACD.對(duì)于①,當(dāng)M=N≤0時(shí),logaM,logaN都沒有意義,故不成立;對(duì)于②,logaM=logaN,則必有M>0,N>0,M=N;對(duì)于③,當(dāng)M,N互為相反數(shù)且不為0時(shí),也有l(wèi)ogaM2=logaN2,但此時(shí)M≠N;對(duì)于④,當(dāng)M=N=0時(shí),logaM2,logaN2都沒有意義,故不成立.6.下列四個(gè)選項(xiàng)中,正確的選項(xiàng)是 ()A.lg(lg10)=0B.lg(lne)=0C.若10=lgx,則x=10D.由log25x=QUOTE,得x=±5【解析】選AB.因?yàn)閘g10=1,所以lg(lg10)=lg1=0,A正確;因?yàn)閘ne=1,所以lg(lne)=lg1=0,B正確;若10=lgx,則x=1010,C錯(cuò)誤;由log25x=QUOTE,得x=QUOTE=5,D錯(cuò)誤.三、填空題(每小題5分,共10分)7.若a=log92,則9a=________,3a+3-a=________.

【解析】a=log92,則9a=QUOTE=2,所以3a=QUOTE,3a+3-a=QUOTE+QUOTE=QUOTE.答案:2QUOTE8.方程4x-2x-6=0的解為________.

【解析】由4x-2x-6=0,得(2x)2-2x-6=0,解得2x=3,或2x=-2(舍去),所以x=log23.答案:x=log23四、解答題(每小題10分,共20分)9.求下列各式的值:(1)2QUOTE.(2)QUOTE+log7343+102lg5.【解析】(1)2QUOTE=(52QUOTE=QUOTE=4.(2)原式=QUOTE+log773+QUOTE=QUOTE+3+25=QUOTE.10.求下列各式中的x值:(1)logx27=QUOTE.(2)log2x=-QUOTE.(3)x=log3QUOTE.【解析】(1)由logx27=QUOTE,可得QUOTE=27,所以x=2QUOTE=(33QUOTE=32=9.(2)由log2x=-QUOTE,可得x=QUOTE,所以x=QUOTE=QUOTE=QUOTE.(3)由x=log3QUOTE,可得x=log33-2=-2.1.對(duì)數(shù)式log(2x-3)(x-1)中實(shí)數(shù)x的取值范圍是________.

【解析】由題意可得QUOTE解得x>QUOTE,且x≠2,所以實(shí)數(shù)x的取值范圍是QUOTE∪(2,+∞).答案:QUOTE∪(2,+∞)2.已知logax=4,logay=5(a>0,且a≠1),求A=QUOTE的值.【解析】由logax=4,得x=a4,由logay=5,得y=a5,