對(duì)數(shù)函數(shù)同步練習(xí) 高一上學(xué)期數(shù)學(xué)人教A版（2019）必修第一冊(cè)

對(duì)數(shù)函數(shù)同步練習(xí)函數(shù)y=log15(?∞,?1)(3,+∞)(?1,0)∪(3,+∞)(?∞,?1)∪(3,+∞)若a=30.5,b=log23c<b<ac<a<bb<c<aa<c<b若2m=3n=p6626若實(shí)數(shù)x,y滿足x>y>0,且log36633已知集合A={x|f(x)=x+2x?5[?2,+∞)(5,+∞)(2,5)∪(5,+∞)(3,5)∪(5,+∞)(多選)若0<m<n<1,則函數(shù)fx=若x2?3x<0若a若a,b為實(shí)數(shù),則2a已知x=log715?log7若函數(shù)fx=3x的反函數(shù)為y=g(x),則函數(shù)fx=1log2已知lgx+lgy=1,則52x+4已知函數(shù)fx=(log3x)2已知函數(shù)f(x)滿足f當(dāng)a=1時(shí),解不等式f若關(guān)于x的方程f(x)=4log19

【答案】對(duì)數(shù)函數(shù)定義域:x2?2x?3>0x>0解得x>3或?1<x<0a=30.5因?yàn)閎=bc=ln?(ln2),因?yàn)閘n1<ln2<lne,故0<ln2<1,所以綜上,c<a<b,故本題B項(xiàng)正確.因?yàn)?m=3n=p,所以又因?yàn)?m+1n=2,所以ln6x2又因?yàn)閤>y>0,故x因此x?由對(duì)數(shù)換底公式可得:log9y故log3x+log所以x?由題意可得:集合A元素是x,即集合A是求解函數(shù)f(x)=x+2即x+2≥0x≠5,解得x∈[?2,5)∪(5,+∞)集合B元素是y,即集合B是求解函數(shù)y=log其中,g(x)=x2?2x+8=x?12+7,故函數(shù)故當(dāng)x>2時(shí),g又因?yàn)閷?duì)數(shù)函數(shù)y的底數(shù)大于1,即對(duì)數(shù)函數(shù)y在定義域內(nèi)單調(diào)遞增,故當(dāng)x>2時(shí),log2x2?2x+8>log28=3,即集合B可進(jìn)一步表示A項(xiàng),對(duì)數(shù)函數(shù)logax過定點(diǎn)(1,0),故令3x?2=1,解得x=1,即函數(shù)fx=log33x?2恒過定點(diǎn)(1,0),所以函數(shù)fx=log33x?2?1的圖像過定點(diǎn)(1,?1),A項(xiàng)錯(cuò)誤;B項(xiàng),因?yàn)閤2?3x<0,解得x∈(0,3),故函數(shù)fx=log3x的值域?yàn)??∞,1),故B項(xiàng)正確;C項(xiàng),loga由題意可得x=log由指數(shù)函數(shù)與對(duì)數(shù)函數(shù)關(guān)系可得7x因?yàn)閒x=3x故g1由題意可得,x?3>0log2x?3>0,解得x>4因?yàn)閤∈(4,+∞),所以x?3∈(1,+∞),log2x?3∈(0,+∞),log2(x?3)∈(0,+∞)由均值不等式可得52x又因?yàn)閘gx+lgy=lgxy=1將xy=10代入可得52x故52x由對(duì)數(shù)換底公式可得:log9設(shè)log3x=t故fx=t因此,當(dāng)t=?14時(shí),1)若a=1,可得fx?log31x2)fx=由對(duì)