人教A版必修一 2.2.1.2对数的运算 课时作业.doc

第2课时对数的运课后篇巩固提升1.2log510+log50.25=()a.0b.1c.2d.4解析:原式=log5102+log50.25=log5(1000.25)=log525=2.答案:c2.若log23=a,则log49=()a.ab.ac.2ad.a2解析:log49=log29log24=2log232=log23=a,故选b.答案:b3.已知4a=7,6b=8,则log1221可以用a,b表示为()a.3-b+2ab3+bb.2a+b-ab3+bc.3-b+2ab4-2bd.2a+b-ab4-2b解析:由题意可得a=log47=lg72lg2,则lg7lg2=2a,b=log68=3lg2lg6,则lg6lg2=3b,据此有:log1221=lg21lg12=lg6+lg7-lg2lg6+lg2=lg6lg2+lg7lg2-1lg6lg2+1=3b+2a-13b+1=3+2ab-b3+b.答案:a4.1log1419+1log1513等于()a.lg 3b.-lg 3c.1lg3d.-1lg3解析:原式=log1914+log1315=log94+log35=log32+log35=log310=1lg3.答案:c5.若2lg(x-2y)=lg x+lg y(x2y0),则yx的值为()a.4b.1或14c.1或4d.14解析:2lg(x-2y)=lg x+lg y(x2y0),lg(x-2y)2=lg xy,(x-2y)2=xy,x2-5xy+4y2=0,(x-y)(x-4y)=0,x=y或x=4y.x-2y0,且x0,y0,xy,yx=14.答案:d6.某种食品因存放不当受到细菌的侵害.据观察,此食品中细菌的个数y与经过的时间t(单位:min)满足关系y=2t,若细菌繁殖到3个,6个,18个所经过的时间分别为t1,t2,t3,则有()a.t1t2=t3b.t1+t2t3c.t1+t2=t3d.t1+t20,0),那么的值是.解析:由题意,得lg +lg =-(lg 7+lg 5)=lg135,所以lg()=lg135,=135.答案:13511.计算:(1)lg2+lg5-lg8lg50-lg40;(2)lg12-lg58+lg54-log92log43.解(1)原式=lg258lg5040=lg54lg54=1.(2)(方法一)原式=lg1258+lg54-lg2lg9lg3lg4=lg4554-lg22lg3lg32lg2=lg 1-14=-14.(方法二)原式=(lg 1-lg 2)-(lg 5-lg 8)+(lg 5-lg 4)-lg2lg9lg3lg4=-lg 2+lg 8-lg 4-lg22lg3lg32lg2=-(lg 2+lg 4)+lg 8-14=-lg(24)+lg 8-14=-14.12.若logax=2,logbx=3,logcx=6,求logabcx的值.解logax=2,logxa=12.同理,logxb=13,logxc=16.logabcx=1logx(abc)=1logxa+logxb+logxc=112+13+16=1.13.导学号03814039已知loga(x2+4)+loga(y2+1)=loga5+loga(2xy-1)(a0,且a1),求log8yx的值.解由对数的运算法则,可将等式化为loga(x2+4)(y2+1) =loga5(2xy-1) ,(x2+4)(y2+1)=5(2xy-1).整理,得x2y2+x2+4y2-10xy+9