PAGEPAGE4第2課時(shí)對(duì)數(shù)的運(yùn)算課時(shí)過關(guān)·實(shí)力提升基礎(chǔ)鞏固1.若a>0,且a≠1,x>y>0,則下列式子正確的個(gè)數(shù)是()①logax·logay=loga(x+y);②logax-logay=loga(x-y);③logaxy=logA.0 B.1 C.2 D.3答案:A2.2log510+log50.25等于()A.0 B.1 C.2 D.4解析:原式=log5100+log50.25=log525=log552=2.答案:C3.計(jì)算log225·log322·log59的結(jié)果為()A.3 B.4 C.5 D.6解析:原式答案:D4.計(jì)算8A.16BCD解析:原式=(23答案:C5.已知log23=a,log37=b,則log27等于()A.a+b B.a-b C.ab D.解析:log27=log23·log37=ab.答案:C6.若lgx-lgy=t,則lgxA.3t B.CD解析:lgx23-lgy23答案:A7.若lgx=lgm-2lgn,則x=.

解析:∵lgm-2lgn=lgm-lgn2=lgm答案:m8.已知3a=2,用a表示log34-log36=.

解析:∵3a=2,∴a=log32,∴l(xiāng)og34-log36=log322-log3(2×3)=2log32-log32-log33=a-1.答案:a-19.若2.5x=1000,0.25y=1000,則1解析:∵x=log2.51000,y=log0.251000,∴1x=同理1y=log10000.25,∴1x-1y=log答案:110.計(jì)算:(1)(lg5)2+3lg2+2lg5+lg2×lg5;(2)(3)(log62)2+(log63)2+3log62×解:(1)(lg5)2+3lg2+2lg5+lg2×lg5=lg5(lg5+lg2)+2(lg2+lg5)+lg2=lg5×lg10+2lg10+lg2=2+(lg5+lg2)=3.(2)(3)(log62)2+(log63)2+3log62×(lo=(log62)2+(log63)2+3log62×lo=(log62)2+(log63)2+3log62×lo=(log62)2+(log63)2+2log62×log63=(log62+log63)2=1.實(shí)力提升1.若lga+lgb=0(其中a>0,b>0,a≠1,b≠1),則函數(shù)f(x)=ax與g(x)=bx的圖象關(guān)于()A.直線y=x對(duì)稱 B.x軸對(duì)稱C.y軸對(duì)稱 D.原點(diǎn)對(duì)稱解析:∵lga+lgb=lg(ab)=0,∴ab=1,∴b=∴g(x)=1ax,故函數(shù)f(x)與g(x答案:C2.若實(shí)數(shù)a,b,c滿意25a=403b=2015c=2018,則下列式子正確的是()A.BC.D解析:由已知,得52a=403b=2015c=2024,得2a=log52024,b=log4032024,c=log20152024,所以12a=log20185,1b=答案:A3.★某種食品因存放不當(dāng)受到細(xì)菌的侵害.據(jù)視察,此食品中細(xì)菌的個(gè)數(shù)y與經(jīng)過的時(shí)間t(單位:min)滿意關(guān)系y=2t.若細(xì)菌繁殖到3個(gè)、6個(gè)、18個(gè)所經(jīng)過的時(shí)間分別是t1,t2,t3,則有()A.t1·t2=t3 B.t1+t2>t3 C.t1+t2=t3 D.t1+t2<t3解析:由題意,得2t1=3,2t2=6,2t3=18,則t1=所以t1+t2=log23+log26=log218=t3.答案:C4.計(jì)算lg25+lg2+lg2·lg5=.

解析:原式=lg5(lg5+lg2)+lg2=lg5+lg2=1.答案:15.已知函數(shù)f(x)=x+1,g(x)=-1解析:f(log23)+g(log62)=log23+1-答案:06.若關(guān)于lgx的方程lg2x+(lg7+lg5)lgx+lg7·lg5=0的兩個(gè)根是lgα,lgβ,則αβ的值是.

解析:由題意,得lgα+lgβ=-(lg7+lg5)=lg135,故lg(αβ)=lg答案:17.已知x,y,z為正數(shù),3x=4y=6z,且2x=py.(1)求p;(2)求證:(1)解設(shè)3x=4y=6z=k(明顯k>0,且k≠1),則x=log3k,y=log4k,z=log6k.由2x=py,得2log3k=plog4k=p·lo∵log3k≠0,∴p=2log34.(2)證明∵∴8.★甲、乙兩人在解關(guān)于x的方程log2x+b+c·logx2=0時(shí),甲寫錯(cuò)了常數(shù)b得兩根為1分析:將方程化為關(guān)于log2x的一元二次方程的形式.先利用一元二次方程的根與系數(shù)的關(guān)系求出b和c,再求出原方程的根.解:由原方程可知x>0,且x≠1.原方程可化為log2x+b+c·1log2x=0,即(log2x)2+blog2x+