20xx高考數(shù)學(xué)文人教a版一輪復(fù)習(xí)學(xué)案：26-對(duì)數(shù)與對(duì)數(shù)函數(shù)-【含解析】(參考版)
2025-04-03 03:15本頁面
【正文】 0==1,即0c1,∴ab0,0ac1,即有abac.∵1a+1b=+==c,即0a+bab=c1,∴aba+b0.綜上,aba+b.對(duì)點(diǎn)訓(xùn)練3(1)B (2)A (1)∵log51log52log55,∴0a12,b==log1215=log25log24=2,∵,∴12c1,∴acb,故選B.(2)∵43a=43log53=log5334=log125811,∴a34.∵43b=43log85=log8354=log5126251,∴b34.∵5584,∴54b=54log85=log84551,∴b45.∵13485,∴54c=54log138=log134851,∴c45.綜上,abc.例4(1)C (2)C (1)因?yàn)閘og12a=log2a,所以f(log2a)+f(log12a)=f(log2a)+f(log2a)=2f(log2a),原不等式變?yōu)?f(log2a)≤2f(1),即f(log2a)≤f(1).又因?yàn)閒(x)是定義在R上的偶函數(shù),且在[0,+∞)內(nèi)單調(diào)遞增,所以|log2a|≤1,即1≤log2a≤1,解得12≤a≤2,故選C.(2)由題意可得a0,log2alog2a或a0,log12(a)log2(a),解得a1或1a.對(duì)點(diǎn)訓(xùn)練4(1)(∞,2)∪0,12 (2){x|4x≤5} (1)由已知條件可知,當(dāng)x∈(∞,0)時(shí),f(x)=log2(x).當(dāng)x∈(0,+∞)時(shí),f(x)1,即為log2x1,解得0x12。當(dāng)x0時(shí),f(x)=loga|x|+1(0a1)(1,1),(1,1),可知應(yīng)選A.(2)函數(shù)y=|log2x|12x的零點(diǎn)個(gè)數(shù)即為方程|log2x|==|log2x|及y=12x的圖象(圖象略),不難得出兩個(gè)函數(shù)的圖象有2個(gè)交點(diǎn),故選C.例3(1)A (2)D (1)∵32a=32log32=log3223=log981,∴a23.∵32b=32log53=log5233=log25271,∴b=23,∴ac.(2)∵0==1,即0a1。log323log3234=3log36+3log332+log36log3238=6+32+log398=.例2(1)C (2)0,22 (1)函數(shù)y=2log4(1x)的定義域?yàn)?∞,1),排除A,B。log55=14.對(duì)點(diǎn)訓(xùn)練1(1)B (2)A (1)因?yàn)閍log34=log34a=2,所以4a=32=9,所以4a=14a=.(2)由題意,f(log354)=f(log3544)=flog323,∵當(dāng)x∈[2,2)時(shí),f(x)=13xx4,且log36∈[2,2),log323∈[2,2),∴f(log36)+f(log354)=13log5[10(332)237log72]=(log33341)≤1},則0a1,由此可知y
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