對(duì)數(shù)與對(duì)數(shù)運(yùn)算第課時(shí)-對(duì)數(shù)的運(yùn)算-課件

2.2.1對(duì)數(shù)與對(duì)數(shù)運(yùn)算第二課時(shí)對(duì)數(shù)的運(yùn)算第二章2.2.1對(duì)數(shù)與對(duì)數(shù)運(yùn)算第二章互動(dòng)課堂2隨堂測(cè)評(píng)3課后強(qiáng)化作業(yè)4預(yù)習(xí)導(dǎo)學(xué)1互動(dòng)課堂2隨堂測(cè)評(píng)3課后強(qiáng)化作業(yè)4預(yù)習(xí)導(dǎo)學(xué)1預(yù)習(xí)導(dǎo)學(xué)預(yù)習(xí)導(dǎo)學(xué)●課標(biāo)展示1．理解對(duì)數(shù)的運(yùn)算性質(zhì)．2．能用換底公式將一般對(duì)數(shù)轉(zhuǎn)化成自然對(duì)數(shù)或常用對(duì)數(shù)．3．了解對(duì)數(shù)在簡(jiǎn)化運(yùn)算中的作用．對(duì)數(shù)與對(duì)數(shù)運(yùn)算第課時(shí)-對(duì)數(shù)的運(yùn)算-課件●溫故知新舊知再現(xiàn)1．對(duì)數(shù)logaN(a>0，且a≠1)具有下列簡(jiǎn)單性質(zhì)：(1)___________沒(méi)有對(duì)數(shù)，即N_____0；(2)1的對(duì)數(shù)為_(kāi)____，即loga1＝_____；(3)底的對(duì)數(shù)等于_____；即logaa＝_____；(4)logaab＝_____.2．對(duì)數(shù)與指數(shù)間的關(guān)系：當(dāng)a>0，a≠1時(shí)，ax＝N?x＝__________.零和負(fù)數(shù)>0011blogaN●溫故知新零和負(fù)數(shù)>0011blogaN3．指數(shù)的運(yùn)算法則：a>0，b>0，r，s∈R，ar·as＝_____，ar÷as＝_____，(ar)s＝_____，(ab)r＝_____.ar＋sar－sarsarbrar＋sar－sarsarbr新知導(dǎo)學(xué)1．對(duì)數(shù)的運(yùn)算性質(zhì)logaM＋logaNlogaM－logaNnlogaM新知導(dǎo)學(xué)logaM＋logaNlogaM－logaNnlog對(duì)數(shù)與對(duì)數(shù)運(yùn)算第課時(shí)-對(duì)數(shù)的運(yùn)算-課件對(duì)數(shù)與對(duì)數(shù)運(yùn)算第課時(shí)-對(duì)數(shù)的運(yùn)算-課件對(duì)數(shù)與對(duì)數(shù)運(yùn)算第課時(shí)-對(duì)數(shù)的運(yùn)算-課件●自我檢測(cè)1．lg2＋lg5的值為()A．2 B．5C．7 D．1[答案]D對(duì)數(shù)與對(duì)數(shù)運(yùn)算第課時(shí)-對(duì)數(shù)的運(yùn)算-課件2．log318－log32的值為()A．log316 B．log320C．log336 D．2[答案]D3．log210·lg4＝________.[答案]24．log29·log278＝________.[答案]2[解析]log29·log278＝log232log3323＝2log23·log32＝2.2．log318－log32的值為()互動(dòng)課堂互動(dòng)課堂1 用logax，logay，logaz表示：對(duì)數(shù)的運(yùn)算性質(zhì)

●典例探究

11 用logax，logay，logaz表示：對(duì)數(shù)的運(yùn)算對(duì)數(shù)與對(duì)數(shù)運(yùn)算第課時(shí)-對(duì)數(shù)的運(yùn)算-課件

規(guī)律總結(jié)：對(duì)于底數(shù)相同的對(duì)數(shù)式的化簡(jiǎn)，常用的方法是：(1)“收”：將同底的兩對(duì)數(shù)的和(差)收成積(商)的對(duì)數(shù)．(2)“拆”：將積(商)的對(duì)數(shù)拆成對(duì)數(shù)的和(差)．(3)對(duì)數(shù)的化簡(jiǎn)求值一般是正用或逆用公式，對(duì)真數(shù)進(jìn)行處理，選哪種策略化簡(jiǎn)，取決于問(wèn)題的實(shí)際情況，一般本著便于真數(shù)化簡(jiǎn)的原則進(jìn)行．

規(guī)律總結(jié)：對(duì)于底數(shù)相同的對(duì)數(shù)式的化簡(jiǎn)，常用的方法是：11對(duì)數(shù)與對(duì)數(shù)運(yùn)算第課時(shí)-對(duì)數(shù)的運(yùn)算-課件2運(yùn)用對(duì)數(shù)的運(yùn)算性質(zhì)解題

22運(yùn)用對(duì)數(shù)的運(yùn)算性質(zhì)解題2[分析]

1.當(dāng)對(duì)數(shù)的底數(shù)相同時(shí)，利用對(duì)數(shù)運(yùn)算的性質(zhì)，將式子轉(zhuǎn)化為只含一種或少數(shù)幾種真數(shù)的形式再進(jìn)行計(jì)算．2．先將45用2與3的冪積表示；再運(yùn)用對(duì)數(shù)的運(yùn)算法則求解．[分析]1.當(dāng)對(duì)數(shù)的底數(shù)相同時(shí)，利用對(duì)數(shù)運(yùn)算的性質(zhì)，將式子對(duì)數(shù)與對(duì)數(shù)運(yùn)算第課時(shí)-對(duì)數(shù)的運(yùn)算-課件對(duì)數(shù)與對(duì)數(shù)運(yùn)算第課時(shí)-對(duì)數(shù)的運(yùn)算-課件對(duì)數(shù)與對(duì)數(shù)運(yùn)算第課時(shí)-對(duì)數(shù)的運(yùn)算-課件

規(guī)律總結(jié)：靈活運(yùn)用對(duì)數(shù)運(yùn)算法則進(jìn)行對(duì)數(shù)運(yùn)算，要注意法則的正用和逆用．在化簡(jiǎn)變形的過(guò)程中，要善于觀察、比較和分析，從而選擇快捷、有效的運(yùn)算方案進(jìn)行對(duì)數(shù)運(yùn)算．

規(guī)律總結(jié)：靈活運(yùn)用對(duì)數(shù)運(yùn)算法則進(jìn)行對(duì)數(shù)運(yùn)算，要注意法則的正22對(duì)數(shù)與對(duì)數(shù)運(yùn)算第課時(shí)-對(duì)數(shù)的運(yùn)算-課件[分析](1)將底統(tǒng)一成以10為底的常用對(duì)數(shù)；(2)等式左邊前一個(gè)對(duì)數(shù)的真數(shù)是后面對(duì)數(shù)的底數(shù)，利用換底公式很容易進(jìn)行約分求解m的值．換底公式的應(yīng)用

[分析](1)將底統(tǒng)一成以10為底的常用對(duì)數(shù)；(2)等式左對(duì)數(shù)與對(duì)數(shù)運(yùn)算第課時(shí)-對(duì)數(shù)的運(yùn)算-課件

規(guī)律總結(jié)：關(guān)于換底公式的用途和本質(zhì)：(1)換底公式的主要用途在于將一般對(duì)數(shù)式化為常用對(duì)數(shù)或自然對(duì)數(shù)，然后查表求值，以此來(lái)解決對(duì)數(shù)求值的問(wèn)題．(2)換底公式的本質(zhì)是化為同底，這是解決對(duì)數(shù)問(wèn)題的基本方法．

規(guī)律總結(jié)：關(guān)于換底公式的用途和本質(zhì)：33對(duì)數(shù)與對(duì)數(shù)運(yùn)算第課時(shí)-對(duì)數(shù)的運(yùn)算-課件對(duì)數(shù)與對(duì)數(shù)運(yùn)算第課時(shí)-對(duì)數(shù)的運(yùn)算-課件對(duì)數(shù)與對(duì)數(shù)運(yùn)算第課時(shí)-對(duì)數(shù)的運(yùn)算-課件[錯(cuò)因分析]在對(duì)數(shù)式的變形過(guò)程中，變形前后字母的取值范圍會(huì)發(fā)生變化，這時(shí)一定要通過(guò)限制條件來(lái)保證變形的等價(jià)性．本題中，去掉對(duì)數(shù)符號(hào)后，x>0，y>0，x－2y>0，這些條件在整式中是體現(xiàn)不出來(lái)的．故應(yīng)添上或在最后進(jìn)行檢驗(yàn)．對(duì)數(shù)與對(duì)數(shù)運(yùn)算第課時(shí)-對(duì)數(shù)的運(yùn)算-課件對(duì)數(shù)與對(duì)數(shù)運(yùn)算第課時(shí)-對(duì)數(shù)的運(yùn)算-課件(2013～2014南陽(yáng)高一檢測(cè))作為對(duì)數(shù)運(yùn)算法則：lg(a＋b)＝lga＋lgb(a＞0，b＞0)是不正確的．但對(duì)一些特殊值是成立的，例如：lg(2＋2)＝lg2＋lg2.那么，對(duì)于所有使lg(a＋b)＝lga＋lgb(a＞0，b＞0)成立的a，b應(yīng)滿足的函數(shù)表達(dá)式a＝f(b)為_(kāi)_______．(2013～2014南陽(yáng)高一檢測(cè))作為對(duì)數(shù)運(yùn)算法則：lg(a對(duì)數(shù)與對(duì)數(shù)運(yùn)算第課時(shí)-對(duì)數(shù)的運(yùn)算-課件隨堂測(cè)評(píng)隨堂測(cè)評(píng)[答案]A[答案]A2．log38·log23＝()A．2 B．3C．4 D．9[答案]B3．已知a＝log32，那么log38－2log36用a表示為()A．a(chǎn)－2 B．5a－2C．3a－(1＋a)2 D．3a－a2－1[答案]A對(duì)數(shù)與對(duì)數(shù)運(yùn)算第課時(shí)-對(duì)數(shù)的運(yùn)算-課件對(duì)數(shù)與對(duì)數(shù)運(yùn)算第課時(shí)-對(duì)數(shù)的運(yùn)算-課件對(duì)數(shù)與對(duì)數(shù)運(yùn)算第課時(shí)-對(duì)數(shù)的運(yùn)算-課件對(duì)數(shù)與對(duì)數(shù)運(yùn)算第課時(shí)-對(duì)數(shù)的運(yùn)算-課件對(duì)數(shù)與對(duì)數(shù)運(yùn)算第課時(shí)-對(duì)數(shù)的運(yùn)算-課件2.2.1對(duì)數(shù)與對(duì)數(shù)運(yùn)算第二課時(shí)對(duì)數(shù)的運(yùn)算第二章2.2.1對(duì)數(shù)與對(duì)數(shù)運(yùn)算第二章互動(dòng)課堂2隨堂測(cè)評(píng)3課后強(qiáng)化作業(yè)4預(yù)習(xí)導(dǎo)學(xué)1互動(dòng)課堂2隨堂測(cè)評(píng)3課后強(qiáng)化作業(yè)4預(yù)習(xí)導(dǎo)學(xué)1預(yù)習(xí)導(dǎo)學(xué)預(yù)習(xí)導(dǎo)學(xué)●課標(biāo)展示1．理解對(duì)數(shù)的運(yùn)算性質(zhì)．2．能用換底公式將一般對(duì)數(shù)轉(zhuǎn)化成自然對(duì)數(shù)或常用對(duì)數(shù)．3．了解對(duì)數(shù)在簡(jiǎn)化運(yùn)算中的作用．對(duì)數(shù)與對(duì)數(shù)運(yùn)算第課時(shí)-對(duì)數(shù)的運(yùn)算-課件●溫故知新舊知再現(xiàn)1．對(duì)數(shù)logaN(a>0，且a≠1)具有下列簡(jiǎn)單性質(zhì)：(1)___________沒(méi)有對(duì)數(shù)，即N_____0；(2)1的對(duì)數(shù)為_(kāi)____，即loga1＝_____；(3)底的對(duì)數(shù)等于_____；即logaa＝_____；(4)logaab＝_____.2．對(duì)數(shù)與指數(shù)間的關(guān)系：當(dāng)a>0，a≠1時(shí)，ax＝N?x＝__________.零和負(fù)數(shù)>0011blogaN●溫故知新零和負(fù)數(shù)>0011blogaN3．指數(shù)的運(yùn)算法則：a>0，b>0，r，s∈R，ar·as＝_____，ar÷as＝_____，(ar)s＝_____，(ab)r＝_____.ar＋sar－sarsarbrar＋sar－sarsarbr新知導(dǎo)學(xué)1．對(duì)數(shù)的運(yùn)算性質(zhì)logaM＋logaNlogaM－logaNnlogaM新知導(dǎo)學(xué)logaM＋logaNlogaM－logaNnlog對(duì)數(shù)與對(duì)數(shù)運(yùn)算第課時(shí)-對(duì)數(shù)的運(yùn)算-課件對(duì)數(shù)與對(duì)數(shù)運(yùn)算第課時(shí)-對(duì)數(shù)的運(yùn)算-課件對(duì)數(shù)與對(duì)數(shù)運(yùn)算第課時(shí)-對(duì)數(shù)的運(yùn)算-課件●自我檢測(cè)1．lg2＋lg5的值為()A．2 B．5C．7 D．1[答案]D對(duì)數(shù)與對(duì)數(shù)運(yùn)算第課時(shí)-對(duì)數(shù)的運(yùn)算-課件2．log318－log32的值為()A．log316 B．log320C．log336 D．2[答案]D3．log210·lg4＝________.[答案]24．log29·log278＝________.[答案]2[解析]log29·log278＝log232log3323＝2log23·log32＝2.2．log318－log32的值為()互動(dòng)課堂互動(dòng)課堂1 用logax，logay，logaz表示：對(duì)數(shù)的運(yùn)算性質(zhì)

●典例探究

11 用logax，logay，logaz表示：對(duì)數(shù)的運(yùn)算對(duì)數(shù)與對(duì)數(shù)運(yùn)算第課時(shí)-對(duì)數(shù)的運(yùn)算-課件

規(guī)律總結(jié)：對(duì)于底數(shù)相同的對(duì)數(shù)式的化簡(jiǎn)，常用的方法是：(1)“收”：將同底的兩對(duì)數(shù)的和(差)收成積(商)的對(duì)數(shù)．(2)“拆”：將積(商)的對(duì)數(shù)拆成對(duì)數(shù)的和(差)．(3)對(duì)數(shù)的化簡(jiǎn)求值一般是正用或逆用公式，對(duì)真數(shù)進(jìn)行處理，選哪種策略化簡(jiǎn)，取決于問(wèn)題的實(shí)際情況，一般本著便于真數(shù)化簡(jiǎn)的原則進(jìn)行．

規(guī)律總結(jié)：對(duì)于底數(shù)相同的對(duì)數(shù)式的化簡(jiǎn)，常用的方法是：11對(duì)數(shù)與對(duì)數(shù)運(yùn)算第課時(shí)-對(duì)數(shù)的運(yùn)算-課件2運(yùn)用對(duì)數(shù)的運(yùn)算性質(zhì)解題

22運(yùn)用對(duì)數(shù)的運(yùn)算性質(zhì)解題2[分析]

1.當(dāng)對(duì)數(shù)的底數(shù)相同時(shí)，利用對(duì)數(shù)運(yùn)算的性質(zhì)，將式子轉(zhuǎn)化為只含一種或少數(shù)幾種真數(shù)的形式再進(jìn)行計(jì)算．2．先將45用2與3的冪積表示；再運(yùn)用對(duì)數(shù)的運(yùn)算法則求解．[分析]1.當(dāng)對(duì)數(shù)的底數(shù)相同時(shí)，利用對(duì)數(shù)運(yùn)算的性質(zhì)，將式子對(duì)數(shù)與對(duì)數(shù)運(yùn)算第課時(shí)-對(duì)數(shù)的運(yùn)算-課件對(duì)數(shù)與對(duì)數(shù)運(yùn)算第課時(shí)-對(duì)數(shù)的運(yùn)算-課件對(duì)數(shù)與對(duì)數(shù)運(yùn)算第課時(shí)-對(duì)數(shù)的運(yùn)算-課件

規(guī)律總結(jié)：靈活運(yùn)用對(duì)數(shù)運(yùn)算法則進(jìn)行對(duì)數(shù)運(yùn)算，要注意法則的正用和逆用．在化簡(jiǎn)變形的過(guò)程中，要善于觀察、比較和分析，從而選擇快捷、有效的運(yùn)算方案進(jìn)行對(duì)數(shù)運(yùn)算．

規(guī)律總結(jié)：靈活運(yùn)用對(duì)數(shù)運(yùn)算法則進(jìn)行對(duì)數(shù)運(yùn)算，要注意法則的正22對(duì)數(shù)與對(duì)數(shù)運(yùn)算第課時(shí)-對(duì)數(shù)的運(yùn)算-課件[分析](1)將底統(tǒng)一成以10為底的常用對(duì)數(shù)；(2)等式左邊前一個(gè)對(duì)數(shù)的真數(shù)是后面對(duì)數(shù)的底數(shù)，利用換底公式很容易進(jìn)行約分求解m的值．換底公式的應(yīng)用

[分析](1)將底統(tǒng)一成以10為底的常用對(duì)數(shù)；(2)等式左對(duì)數(shù)與對(duì)數(shù)運(yùn)算第課時(shí)-對(duì)數(shù)的運(yùn)算-課件

規(guī)律總結(jié)：關(guān)于換底公式的用途和本質(zhì)：(1)換底公式的主要用途在于將一般對(duì)數(shù)式化為常用對(duì)數(shù)或自然對(duì)數(shù)，然后查表求值，以此來(lái)解決對(duì)數(shù)求值的問(wèn)題．(2)換底公式的本質(zhì)是化為同底，這是解決對(duì)數(shù)問(wèn)題的基本方法．

規(guī)律總結(jié)：關(guān)于換底公式的用途和本質(zhì)：33對(duì)數(shù)與對(duì)數(shù)運(yùn)算第課時(shí)-對(duì)數(shù)的運(yùn)算-課件對(duì)數(shù)與對(duì)數(shù)運(yùn)算第課時(shí)-對(duì)數(shù)的運(yùn)算-課件對(duì)數(shù)與對(duì)數(shù)運(yùn)算第課時(shí)-對(duì)數(shù)的運(yùn)算-課件[錯(cuò)因分析]在對(duì)數(shù)式的變形過(guò)程中，變形前后字母的取值范圍會(huì)發(fā)生變化，這時(shí)一定要通過(guò)限制條件來(lái)保證變形的等價(jià)性．本題中，去掉對(duì)數(shù)符號(hào)后，x>0，y>0，x－2y>0，這些條件在整式中是體現(xiàn)不出來(lái)的．故應(yīng)添上或在最后進(jìn)行檢驗(yàn)．對(duì)數(shù)與對(duì)數(shù)運(yùn)算第課時(shí)-對(duì)數(shù)的運(yùn)算-課件對(duì)數(shù)與對(duì)數(shù)運(yùn)算第課時(shí)-對(duì)數(shù)的運(yùn)算-課件(2013～2014南陽(yáng)高一檢測(cè))作為對(duì)數(shù)運(yùn)算法則：lg(a＋b)＝lga＋lgb(a＞0，b＞0)是不正確的．但對(duì)一些特殊值是成立的，例如：lg(2＋2)＝lg2＋lg2.那么，對(duì)于所有使lg(a＋b)＝lga＋lgb(a＞0，b＞0)成立的a，b應(yīng)滿足的函數(shù)表達(dá)式a＝f(b)為_(kāi)_______．(2013～2014南陽(yáng)高一檢測(cè))作為對(duì)數(shù)運(yùn)算法則：lg(a對(duì)數(shù)與對(duì)數(shù)運(yùn)算第課時(shí)-對(duì)數(shù)的運(yùn)算-課件隨堂測(cè)評(píng)