2017-2018學年高中數(shù)學課時作業(yè)262.2.2對數(shù)與對數(shù)運算(第2課時)新必

1、A.lg2 lg3B.lg2 + lg3課時作業(yè)（二十六）221.2對數(shù)與對數(shù)運算（第 2 課時）1.log35 log345=()C.6D. 622B. 1D. 2A.1C.2答案2.若 lgx = lga + 2lgb 3lgc，貝 U x =(A.a + 2b 3c2abB.3cab2C.3-cD.ab答案3.當 a0,1 時，下列說法正確的是若 M= N 貝UlogaMklogaN;若若logaM = logaN2,則Mk N;若logaM=logaN,貝UM=A.與B.與C.D.答案 C4 g(100 x)A.200C.4答案 C4B.105.已知 |lga| = lgb(a0 ,

2、b0)，那么(A.a = bB.a = b 或 ab= 1C.a = bD.ab = 1答案 B6.已知 2log6x= 1 log63,則 x 的值是(A. 3B. ,2C2 或一. 2D. .3 或 . 2答案 B27.設(shè)x比lg100 大(C.6D. 623答案解析12.(1)2log510+ log50.25 = 答案2149 + log213 log217 = 答案答案 C解析 設(shè) Igx = t，則 t + (Ig2 + Ig3)t + Ig2lg3 = 0.,11= lgx1,1據(jù) v又 11+ 12= lg2 lg3 = lgx1+ Igx2,. X1X2= .12= lgx2

3、,68.已知 log32= a, log35= b，貝Ulog310 等于()A.a + bB.a bC.abaD-b答案 A解析 log310= log3(2x5) = log32+ log35.b9.已知 lga = 2.431 0 , lgb = 1.431 0，則-等于(aC.10答案解析1.431b 10-=.n2.431= 10a 101=1，故選 B. I10.已知 2x=A.log215C.log325答案B.logD.log解析11.log523310A由已知 x = log23, x + y = log23 + log25 = log215.232 邁邁log2p2= =_

4、.(2)log原式=log=log232= 5.449X3解析 原式=log217 = 1.(3)lg75 lg5 lg3 + lg2 =_答案 175X2解析 原式=lg 5x3=1.5113._ 求值：lg2.5 lg 石 + lg 匸=82答案 lg2214._ (1)若 lg2 = a, lg3 = b,則 lg =答案a b解析 原式=lg2 lg3 = a b.121332)2+ log0.254+ 9log55 log 31 =1211923原式=(2)+ log0.25O.25 + 9log552 0=才才 + 1 + 2=匸，x3y315._ 若 ln x ln y = a，

5、則 ln( 3) ln(專專) )等于_答案 3a16. 計算.3(1) lg 7 + lg70 lg3 ；(2) lg22 + lg5lg20 1;222Ig52+ ?lg8 + lg5 lg20 + (lg2)2.答案(1)1(2)0(3)32 2解析 (3)原式=2(lg5 + lg2) + lg5(lg5 + 2lg2) + (lg2)= 2+ (lg5 + lg2) = 2+ 1 = 3.17. 若 lga , lgb 是方程 2x2 4x + 1 = 0 的兩個根，則(lg 7)2的值等于()b1A.2B.21C.4D.-4答案 A(2)(log答案23T解析51a222解析 / lga + lgb = 2, lga Igb = ，二(lg R=(lga lgb) = (Iga + Igb) 4lga Igb =2.?重點班選做題18.已知 loga2= m, loga3= n.(1) 求 a2mn的值； (2)求 loga18.解析 (1) .Toga2= m, loga3= n,. O = 2, a = 3.2mn2m n . m 2 n_4a=a-a=(aa=2-3=3 2 2(2) loga18=loga(2X3)=loga2+loga3=loga2 + 2loga3= nn