1、4对数第1课时对数及其运算基础过关1将23化为对数式为()Alog23 Blog(3)2Clog23 Dlog2(3)解析根据对数的定义知选C.答案C2有以下四个结论:lg(lg 10)0;ln(ln e)0;若10lg x,则x10;若eln x,则xe2.其中正确的是()A B C D解析lg(lg 10)lg 10,ln(ln e)ln 10,故正确;若10lg x,则x1010,故错误;若eln x,则xee,故错误答案C3若log3(log2x)1,则x()A. B. C. D.解析log3(log2x)1,log2x3,x238,则x.答案C4ln 1log(1)(1)_解析ln
2、1log(1)(1)011.答案15方程9x63x70的解是_解析设3xt(t0),则原方程可化为t26t70,解得t7或t1(舍去),t7,即3x7.xlog37.答案xlog376求下列各式中的x的值:(1)logx27;(2)log2x;(3)logx(32)2;(4)log5(log2x)0;(5)xlog27.解(1)由logx27得x27,x27329.(2)由log2x得2x,x.(3)由logx(32)2得32x2,即x(32)1.(4)由log5(log2x)0得log2x1.x212.(5)由xlog27得27x,即33x32,x.7(1)若f(10x)x,求f(3)的值;
3、(2)计算23log2335log39.解(1)令t10x,则xlg t,f(t)lg t,即f(x)lg x,f(3)lg 3.(2)23log2335log39232log23233242751.能力提升8方程2log3x的解是()Ax Bx Cx Dx9解析2log3x22,log3x2,x32.答案A9已知loga2m,loga3n,则a2mn()A5 B7 C10 D12解析am2,an3,a2mna2man(am)2an12.答案D10若log(1x)(1x)21,则x_解析由题意知1x(1x)2,解得x0,或x3,验证知,当x0时,log(1x)(1x)2无意义,当x0时不合题意
4、,应舍去x3.答案311若alg 2,blg 3,则100a的值为_解析alg 2,10a2.blg 3,10b3.100a.答案12已知log2(log3(log4x)0,且log4(log2y)1.求y的值解log2(log3(log4x)0,log3(log4x)1,即log4x3,x4364,同理y16,y1682364.创新突破13已知log2(log(log2x)log3(log(log3y)log5(log(log5z)0,试比较x,y,z的大小解由log2(log(log2x)0得,log(log2x)1,log2x,即x2;由log3(log(log3y)0得,log(log3y)1,log3y,即y3;由log5(log(log5z)0得,log(log5z)1,log5z,即z5.y339,x228,yx,又x2232,z5525,xz,故yxz.