1知識(shí)梳理·雙基自測(cè)2考點(diǎn)突破·互動(dòng)探究3名師講壇·素養(yǎng)提升1知識(shí)梳理·雙基自測(cè)知識(shí)點(diǎn)一對(duì)數(shù)與對(duì)數(shù)運(yùn)算1．對(duì)數(shù)的概念(1)對(duì)數(shù)的定義：如果ax＝N(a＞0，且a≠1)，那么數(shù)x叫做以a為底N的對(duì)數(shù)，記作_________，其中_________叫做對(duì)數(shù)的底數(shù)，_________叫做真數(shù)．x＝logaNaN(2)幾種常見(jiàn)對(duì)數(shù)logaN對(duì)數(shù)形式特點(diǎn)記法一般對(duì)數(shù)底數(shù)為a(a>0，且a≠1)_________常用對(duì)數(shù)底數(shù)為_(kāi)_________________自然對(duì)數(shù)底數(shù)為_(kāi)_________________10lgNelnN01(其中a>0且a≠1)NlogaM＋logaNlogaM－logaNnlogaM知識(shí)點(diǎn)二對(duì)數(shù)函數(shù)的圖象與性質(zhì)1．對(duì)數(shù)函數(shù)的定義、圖象和性質(zhì)(0，＋∞)(－∞，＋∞)(1，0)y＞0y＜0增函數(shù)

減函數(shù)

2．反函數(shù)指數(shù)函數(shù)y＝ax(a>0且a≠1)與對(duì)數(shù)函數(shù)_________(a>0且a≠1)互為反函數(shù)，它們的圖象關(guān)于直線(xiàn)_________對(duì)稱(chēng)．y＝logaxy＝x1．指數(shù)式與對(duì)數(shù)式互化××××√×0－143．(必修1P74AT4改編)若lg2＝a，lg3＝b，則lg12的值為(

)A．a(chǎn) B．bC．2a＋b D．2ab[解析]

因?yàn)閘g2＝a，lg3＝b，所以lg12＝lg(4×3)＝2lg2＋lg3＝2a＋b.故選C.C5．(必修1P75AT10改編)已知圖中曲線(xiàn)C1，C2，C3，C4是函數(shù)y＝logax的圖象，則曲線(xiàn)C1，C2，C3，C4對(duì)應(yīng)的a的值依次為

(

)BA7．(2017·全國(guó)卷Ⅱ，5分)函數(shù)f(x)＝ln(x2－2x－8)的單調(diào)遞增區(qū)間是

(

)A．(－∞，－2) B．(－∞，1)C．(1，＋∞) D．(4，＋∞)[解析]

由x2－2x－8>0，得x<－2或x>4.因此，函數(shù)f(x)＝ln(x2－2x－8)的定義域是(－∞，－2)∪(4，＋∞)．注意到函數(shù)y＝x2－2x－8在(4，＋∞)上單調(diào)遞增，由復(fù)合函數(shù)的單調(diào)性知，f(x)＝ln(x2－2x－8)的單調(diào)遞增區(qū)間是(4，＋∞)，選D.D2考點(diǎn)突破·互動(dòng)探究考點(diǎn)一對(duì)數(shù)與對(duì)數(shù)運(yùn)算——自主練透例1考點(diǎn)二對(duì)數(shù)函數(shù)的圖象與性質(zhì)D考向1

對(duì)數(shù)函數(shù)的圖象及其應(yīng)用——師生共研例2B應(yīng)用對(duì)數(shù)型函數(shù)的圖象可求解的問(wèn)題(1)對(duì)一些可通過(guò)平移、對(duì)稱(chēng)變換作出其圖象的對(duì)數(shù)型函數(shù)，在求解其單調(diào)性(單調(diào)區(qū)間)、值域(最值)、零點(diǎn)時(shí)，常利用數(shù)形結(jié)合思想．(2)一些對(duì)數(shù)型方程、不等式問(wèn)題常轉(zhuǎn)化為相應(yīng)的函數(shù)圖象問(wèn)題，利用數(shù)形結(jié)合法求解．〔變式訓(xùn)練1〕(1)函數(shù)f(x)＝loga|x|＋1(0<a<1)的圖象大致為

(

)A考向2對(duì)數(shù)函數(shù)的性質(zhì)及其應(yīng)用——多維探究角度1比較對(duì)數(shù)值的大小A例3角度2利用對(duì)數(shù)函數(shù)單調(diào)性求參數(shù)的取值范圍D例4(2021·華南師大附中模擬)已知函數(shù)f(x)＝log0.5(x2－ax＋3a)在[2，＋∞)上單調(diào)遞減，則實(shí)數(shù)a的取值范圍是

(

)A．(－∞，4] B．[4，＋∞)C．[－4，4] D．(－4，4][分析]

函數(shù)f(x)＝log0.5(x2－ax＋3a)在[2，＋∞)上單調(diào)遞減，說(shuō)明在[2，＋∞)上，函數(shù)t＝x2－ax＋3a>0成立，且為增函數(shù)．角度3簡(jiǎn)單對(duì)數(shù)不等式的解法C例51．比較對(duì)數(shù)式的大小的關(guān)系：(1)若底數(shù)為同一常數(shù)，則可由對(duì)數(shù)函數(shù)的單調(diào)性直接進(jìn)行判斷；若底數(shù)為同一字母，則需要對(duì)底數(shù)