f(x)=bf(x)=logab, logaf(x)=bf(x)=a; (定义法) =a=b
g(x)
b
f(x)
f(x)=g(x), logaf(x)=logag(x)f(x)=g(x)>0(转化法) f(x)logma=g(x)logmb (取对数法)
f(x)g(x)
(4) logaf(x)=logbg(x)logaf(x)=logag(x)/logab(换底法) 三、考点逐个突破 1.指数、对数的运算法则
例1 .计算:(1)(124223)27162(8);
(2)(lg2)lg2lg50lg25;
(3)(log32log92)(log43log83) 解:(1)原式(113)212212163423133316243423282(1)3
1133222212311338811
2 (2)原式(lg2)(1lg5)lg2lg5(lg2lg51)lg22lg5 (11)lg22lg52(lg2lg5)2
(3)原式(lg2lg2lg3lg3lg2lg2lg3lg3)()()() lg3lg9lg4lg8lg32lg32lg23lg2 3lg25lg35 2lg36lg24例2. 设、b、为正数,且满足a2b2c2
bcac)log2(1)1 abbc2 (2)若log4(1)1,log8(abc),求、b、的值
a3abcabcabcabc证明:(1)左边log2log2log2()
abab (1)求证:log2(1(ab)2c2a22abb2c22abc2c2log2log2log2log221;
ababab解:(2)由log4(1bcbc)1得14, aa∴3abc0……………①
2由log8(abc)得abc834………… ……………②
3由①②得ba2……………………………………③ 由①得c3ab,代入a2b2c2得2a(4a3b)0, ∵a0, ∴4a3b0………………………………④ 由③、④解得a6,b8,从而c10 例3. 已知xx1212121223,求
12x2x22xx3232的值
3解:∵xx3,∴(xx)9,∴x2x19,∴xx17,
122 ∴(xx)49,∴x2x247, 又∵xx ∴
323212(xx)(x1x1)3(71)18, 4723
1831212x2x22xx323232.指对互化
例4.已知3a5bc,且
112,求的值 ab1; a解:由3ac得:logc3a1,即alogc31,∴logc3 同理可得
111logc5,∴由2 得 logc3logc52, bab∴logc152,∴c215,∵c0,∴c15 3.换底公式
例5.设x1,y1,且2logxy2logyx30,求Tx24y2的最小值 解:令 tlogxy,∵x1,y1,∴ t0 由2logxy2logyx30得2t230,∴2t23t20, t111 ∴(2t1)(t2)0,∵t0,∴t,即logxy,∴yx2,
22 ∴Tx4yx4x(x2)4, ∵x1,∴当x2时,Tmin4 4.比较大小
例6.(1)若a2ba1,则logb2222b,logba,logab从小到大依次为 ; a (2)若2x3y5z,且,,都是正数,则2x,3y,5z从小到大依次为 ; (3)设x0,且axbx1(a0,b0),则与b的大小关系是( ) A.ba1 B。ab1 C。1ba D。1ab 解:(1)由a2ba1得
bba,故logblogba1logab aa (2)令2x3y5zt,则t1,xlgtlgtlgt,y,z, lg2lg3lg5 ∴2x3y2lgt3lgtlgt(lg9lg8)0,∴2x3y; lg2lg3lg2lg3同理可得:2x5z0,∴2x5z,∴3y2x5z (3)取x1,知选 5.指数函数综合运用 例7.已知函数f(x)axx2(a1), x1求证:(1)函数f(x)在(1,)上为增函数;
(2)方程f(x)0没有负数根 证明:(1)设1x1x2,
则f(x1)f(x2)ax1x12x2 ax22x11x21ax1ax2x12x223(x1x2), ax1ax2x11x21(x11)(x21)∵1x1x2,∴x110,x210,x1x20, ∴
3(x1x2)0;
(x11)(x21)∵1x1x2,且a1,∴ax1ax2,∴ax1ax20, ∴f(x1)f(x2)0,即f(x1)f(x2), ∴函数f(x)在(1,)上为增函数; 另法:∵a1,x(1,) ∴f(x)(axx23)axlna0 2x1(x1)∴函数f(x)在(1,)上为增函数;
(2)假设x0是方程f(x)0的负数根,且x01,则a0xx020, x01 即ax02x03(x01)31, ① x01x01x01333,∴12, x01x01当1x00时,0x011,∴
x0而由a1知a1 ∴①式不成立;
当x01时,x010,∴∴①式不成立
330,∴11,而ax00 x01x01综上所述,方程f(x)0没有负数根 6.对数函数综合运用
例8.已知函数f(x)loga(ax1)(a0且a1)
求证:(1)函数f(x)的图象在轴的一侧;
(2)函数f(x)图象上任意两点连线的斜率都大于0 证明:(1)由ax10得:ax1,
∴当a1时,x0,即函数f(x)的定义域为(0,),此时函数f(x)的图象在轴的右侧; 当0a1时,x0,即函数f(x)的定义域为(,0),此时函数f(x)的图象在轴的左侧 ∴函数f(x)的图象在轴的一侧;
(2)设A(x1,y1)、B(x2,y2)是函数f(x)图象上任意两点,且x1x2, 则直线AB的斜率ky1y2,
x1x2x2y1y2loga(a1)loga(a1)logax1ax11a1xx2,
xxx当a1时,由(1)知0x1x2,∴1a1a2,∴0a11a21,
ax11∴0x1,∴y1y20,又x1x20,∴k0;
a21当0a1时,由(1)知x1x20,∴a1a∴a11a210,
xxxx21,
ax11∴x1,∴y1y20,又x1x20,∴k0 a21∴函数f(x)图象上任意两点连线的斜率都大于0