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)4.2.2對(duì)數(shù)運(yùn)算法則課時(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)算法則(15分鐘30分)1.(2024·全國Ⅰ卷)設(shè)alog34=2,則4-a= ()A.QUOTE B.QUOTE C.QUOTE D.QUOTE【解析】選B.由alog34=2可得log34a=2,所以4a=9,所以有4-a=QUOTE.【補(bǔ)償訓(xùn)練】(2024·合肥高一檢測(cè))lg2·log210的值為 ()A.-1B.0C.1D.2【解析】選C.lg2·log210=lg2·QUOTE=1.2.lgQUOTE= ()A.-4 B.4 C.10 D.-10【解析】選A.lgQUOTE=lg10-4=-4.3.若lgx=m,lgy=n,則lgQUOTE-lgQUOTE的值為 ()A.QUOTEm-2n-2 B.QUOTEm-2n-1C.QUOTEm-2n+1 D.QUOTEm-2n+2【解析】選D.因?yàn)閘gx=m,lgy=n,所以lgQUOTE-lgQUOTE=QUOTElgx-2lgy+2=QUOTEm-2n+2.4.實(shí)數(shù)a,b滿意2a=5b=10,則下列關(guān)系正確的是 ()A.QUOTE+QUOTE=2 B.QUOTE+QUOTE=1C.QUOTE+QUOTE=2 D.QUOTE+QUOTE=QUOTE【解析】選B.因?yàn)?a=5b=10,所以a=log210,b=log510,所以QUOTE=lg2,QUOTE=lg5,所以QUOTE+QUOTE=lg2+lg5=lg(2×5)=1.5.計(jì)算log525+lnQUOTE-(0.64QUOTE=________.

【解析】原式=2+QUOTE-QUOTE=QUOTE.答案:QUOTE6.(1)計(jì)算:(lg2)2+(lg2+3)lg5+lg4.(2)已知log53=a,log54=b,用a,b表示log25144.【解析】(1)原式=(lg2)2+lg2lg5+3lg5+lg4=lg2(lg2+lg5)+lg5+2(lg2+lg5)=lg2+lg5+2=3.(2)因?yàn)閘og53=a,log54=b,所以log25144=log512=log53+log54=a+b.(30分鐘60分)一、單選題(每小題5分,共20分)1.(2024·北京高一檢測(cè))若lga-2lg2=1,則a= ()A.4 B.10 C.20 D.40【解析】選D.lga-2lg2=1化為lga-lg22=1,即lgQUOTE=1,所以,QUOTE=10,a=40.2.已知3x=5y=a,且QUOTE+QUOTE=2,則a的值為 ()A.QUOTE B.15 C.±QUOTE D.225【解析】選A.因?yàn)?x=5y=a所以xlg3=ylg5=lga,所以QUOTE=QUOTE,QUOTE=QUOTE,則2=QUOTE+QUOTE=QUOTE=QUOTE,所以lga2=lg15,因?yàn)閍>0所以a=QUOTE.3.已知實(shí)數(shù)a,b滿意ab=ba,且logab=2,則ab= ()A.QUOTE B.2 C.4 D.8【解析】選D.因?yàn)閷?shí)數(shù)a,b滿意logab=2,故a2=b,又由ab=ba得QUOTE=a2a,解得:a=2,或a=0(舍去),故b=4,ab=8.4.已知2×9x-28=QUOTE,則x= ()A.log37-log32 B.loQUOTE4C.2log32 D.log37【解析】選C.已知2×9x-28=QUOTE,所以2×(3x)2-28-3x=0,即(3x-4)(2·3x+7)=0,解得3x=4,則x=log34=2log32.二、多選題(每小題5分,共10分,全部選對(duì)得5分,選對(duì)但不全的得3分,有選錯(cuò)的得0分)5.已知x,y為正實(shí)數(shù),則 ()A.2lnx+lny=2lnx+2lny B.2ln(x+y)=2lnx·2lnyC.2lnx·lny=(2lnx)lny D.2ln(xy)=2lnx·2lny【解析】選CD.依據(jù)指數(shù)與對(duì)數(shù)的運(yùn)算性質(zhì)可得2lnx·lny=(2lnx)lny,2ln(xy)=2lnx+lny=2lnx·2lny,可知:C,D正確,而A,B都不正確.6.若ab>0,有下列四個(gè)等式:①lg(ab)=lga+lgb;②lgQUOTE=lga-lgb;③QUOTElgQUOTE=lgQUOTE;④lg(ab)=QUOTE,其中不正確的是 ()A.① B.② C.③ D.④【解析】選ABD.①②式成立的前提是a>0,b>0;④式成立的前提是ab≠1.只有③式成立.三、填空題(每小題5分,共10分)7.lgQUOTE+3lg2-QUOTE+e0=________.

【解析】lgQUOTE+3lg2-QUOTE+e0=lgQUOTE-2+1=1.答案:1【補(bǔ)償訓(xùn)練】QUOTE+log2(47×25)-πl(wèi)nQUOTE=________.

【解析】QUOTE+log2(47×25)-πl(wèi)nQUOTE=4-π+log2219+π=4+19=23.答案:238.已知logab+3logba=QUOTE,則logab=________,當(dāng)a>b>1時(shí),QUOTE的值為_______.

【解析】因?yàn)閘ogab+3logba=QUOTE,所以logab+QUOTE=QUOTE,所以2(logab)2-13logab+6=0,解得logab=6或QUOTE,因?yàn)閍>b>1,所以0<logab<1,所以logab=QUOTE,所以QUOTE=b,所以QUOTE=QUOTE=1.答案:6或QUOTE1四、解答題(每小題10分,共20分)9.已知2x=3y=5z,且QUOTE+QUOTE+QUOTE=1,求x,y,z.【解析】令2x=3y=5z=k(k>0),所以x=log2k,y=log3k,z=log5k,所以QUOTE=logk2,QUOTE=logk3,QUOTE=logk5,由QUOTE+QUOTE+QUOTE=1,得logk2+logk3+logk5=logk30=1,所以k=30,所以x=log230=1+log215,y=log330=1+log310,z=log530=1+log56.10.(1)計(jì)算:log3QUOTE+lg5+QUOTE+log23·log94+lg2;(2)若a,b分別是方程(lgx)2-lgx2+QUOTE=0的兩個(gè)實(shí)根,求lgQUOTE·QUOTE的值.【解析】(1)原式=log3QUOTE+lg5+2+QUOTE·QUOTE+lg2=QUOTE-1+QUOTE+3=-QUOTE+1+3=QUOTE.(2)依據(jù)題意,lga,lgb是方程t2-2t+QUOTE=0的兩個(gè)實(shí)根,則lga+lgb=2,lga·lgb=QUOTE,原式=QUOTE=QUOTE·QUOTE=QUOTE·QUOTE=2×QUOTE=12.1.方程log2(x+2)=1+log4(6-x)的解為x=________.

【解析】由log2(x+2)=1+log4(6-x),得:log2(x+2)=log2(2QUOTE),所以x+2=2QUOTE,解得x=2或-10(舍去).答案:22.若a,b,c∈N*,且滿意a2+b2=c2.(1)求log2QUOTE+log2QUOTE的值.(2)若log4QUOTE=1,log8(a+b-c)=QUOTE,求a,b,c的值.【解析】(1)因?yàn)閍2+b2=c2,所以log2QUOTE+log2QUOTE=log2QUOTE=log2QUOTE=log2QUOTE=log2QUOTE=1.(2)因?yàn)閘og4QUOTE=1,所以QUOTE=4.即3a-b-c=0.①因?yàn)閘og8(a+b-c)=QUOTE,所以a+b-c=4.②因?yàn)閍2+b2=c2,③且a,b,c∈N*,所以由①②③解得a=6,b=8,c=10.【補(bǔ)償訓(xùn)練】已知二次函數(shù)f(x)=QUOTEx2+2x+4lga的最小值為3,求QUOTE+loga2·loga50的值.【解析】因?yàn)閒(x)=QUOTEx2+2x+4lga存在最小值為3,所以lga>0,fQUOTE=fQUOTE=lga×QUOTE+