學年高中數(shù)學第二章基本初等函數(shù)Ⅰ22對數(shù)函數(shù)221對數(shù)對數(shù)運算第2課時對數(shù)運算課后篇范文鞏固提升含解析新人教A版必修1

優(yōu)選文檔優(yōu)選文檔PAGEPAGE7優(yōu)選文檔PAGE第2課時對數(shù)的運算課后篇牢固提升基礎(chǔ)牢固1.已知logx16=2,則x等于()A.±4B.4剖析∵logx16=2,∴x2=16.∵x>0且x≠1,∴x=4.答案B2.2log510+log50.25=()剖析原式=log5102+log50.25=log5(100×0.25)=log525=2.答案C3.若log23=a,則log49=()剖析log49==log23=a,應選B.答案B4.等于()剖析原式=lo+lo=log94+log35=log32+log35=log310=.答案C5.若2lg(x-2y)=lgx+lgy(x>2y>0),則的值為()A.4B.1或C.1或4D.剖析∵2lg(x-2y)=lgx+lgy(x>2y>0),∴l(xiāng)g(x-2y)2=lgxy,∴(x-2y)2=xy,∴x2-5xy+4y2=0,∴(x-y)(x-4y)=0,∴x=y或x=4y.∵x-2y>0,且x>0,y>0,∴x≠y,∴.答案D6.計算:2+lg4+2lg5-eln3=.?剖析由題意得2+lg4+2lg5-eln3=(33+(lg4+lg25)-eln3=3+2-3=2.答案27.log35log46log57log68log79=.?剖析log35log46log57log68log79==3.答案38.若2x=3,log4=y,則x+2y=.?剖析∵2x=3,∴x=log23.∴x+2y=log23+2log4=log23+2×=log23+log2=log28=3.答案39.若是關(guān)于lgx的方程lg2x+(lg7+lg5)lgx+lg7·lg5=0的兩個根是lgα,lgβ(α>0,β>0),那么αβ的值是.?剖析由題意,得lgα+lgβ=-(lg7+lg5)=lg,因此lg(αβ)=lg,∴αβ=.答案10.計算:(1);(2)lg-lg+lg-log92·log43.解(1)原式==1.(2)(方法一)原式=lg+lg=lg=lg1-=-.(方法二)原式=(lg1-lg2)-(lg5-lg8)+(lg5-lg4)-=-lg2+lg8-lg4-=-(lg2+lg4)+lg8-=-lg(2×4)+lg8-=-.11.已知log2(log3(log4x))=0,且log4(log2y)=1.求的值.解∵log2(log3(log4x))=0,∴l(xiāng)og3(log4x)=1.∴l(xiāng)og4x=3.∴x=43=64.由log4(log2y)=1,知log2y=4,∴y=24=16.因此×1=8×8=64.能力提升1.若lgx-lgy=a,則lg-lg=()C.aD.剖析lg-lg=3=3(lgx-lgy)=3a.答案A2.若2loga(P-2Q)=logaP+logaQ(a>0,且a≠1),則的值為()C.1D.4或1剖析由2loga(P-2Q)=logaP+logaQ,得loga(P-2Q)2=loga(PQ).由對數(shù)運算法規(guī)得(P-2Q)2=PQ,即P2-5PQ+4Q2=0,因此P=Q(舍去)或P=4Q,解得=4.答案B3.已知0<a<1,x=loga+loga,y=loga5,z=loga-loga,則()A.x>y>zB.z>y>xC.z>x>yD.y>x>z剖析由題意得x=loga+loga=loga,y=loga5=loga,z=loga-loga=loga,因為0<a<1,又,因此loga>loga>loga,即y>x>z,應選D.答案D4.某種食品因存放不當碰到細菌的傷害.據(jù)觀察,此食品中細菌的個數(shù)y與經(jīng)過的時間t(單位:min)滿足關(guān)系y=2t,若細菌生殖到3個,6個,18個所經(jīng)過的時間分別為t1,t2,t3,則有()A.t1·t2=t3B.t1+t2>t3C.t1+t2=t3D.t1+t2<t3剖析由題意,得=3,=6,=18,則t1=log23,t2=log26,t3=log218,因此t1+t2=log23+log26=log218=t3.答案C5.2x=5y=m(m>0),且=2,則m的值為.?剖析由2x=5y=m(m>0),得x=log2m,y=log5m,由=2,得=2,即logm2+logm5=2,logm(2×5)=2.故有m=.答案6.已知a>b>1,若logab+logba=,ab=ba,則a=,b=.?剖析先求出對數(shù)值,再利用指數(shù)相等列方程求解.∵logab+logba=logab+,∴l(xiāng)ogab=2或logab=.∵a>b>1,∴l(xiāng)ogab<logaa=1.∴l(xiāng)ogab=,∴a=b2.∵ab=ba,∴(b2)b=,∴b2b=.∴2b=b2,∴b=2,∴a=4.答案427.已知,log74=b,用a,b表示log4948為.?剖析由可得a=log73,由log74=b可得b=2log72,因此log4948=(4log72+log73)=.答案8.設x,y,z均為正數(shù),且3x=4y=6z,試求x,y,z之間的關(guān)系.解設3x=4y=6z=t,由x>0,知t>1,故取以t為底的對數(shù),可得xlogt3=ylogt4=zlogt6=1,∴x=,y=,z=.∵=logt6-logt3=logt2=logt4=,∴x,y,z之間的關(guān)系為.9.已知loga(x2+4)+loga(y2+1)=loga5+loga(2xy-1)(a>0,且a≠1),求log8的值.解由對數(shù)的運算法規(guī),可將等式化為loga[(x2+4)·(y2+1)]=