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 <tbody> <tr> <td id="artContent" style="max-width: 656px;"> <div style="width: 656px; margin: 0; padding: 0; height: 0;"></div> <p><span><span>一、底数不统一</span></span></p><p><span>对数的运算性质是建立在底数相同的基础上的，但实际问题中，却经常要遇到底数不相同的情况，碰到这种情形，主要有三种处理的方法：</span></p><p><span>1、<span>化为指数式</span></span></p><p><span><span>对数函数与指数函数互为反函数，它们之间有着密切的关系：</span>log<sub>a</sub>N=b<sub><img doc360img-src='http://image109.360doc.com/DownloadImg/2019/06/2608/_1_' data-ratio='0.57143' data-type='gif' data-w='21' _width='-30px' src='http://image109.360doc.com/DownloadImg/2019/06/2608/_1_'></sub>a<sup>b</sup>=N<span>，因此在处理有关对数问题时，经常将对数式化为指数式来帮助解决。</span></span></p><p><span>2、<span>利用换底公式统一底数</span></span></p><p><span>换底公式可以将底数不同的对数通过换底把底数统一起来，然后再利用同底对数相关的性质求解。</span></p><p><span>3、<span>利用函数图象</span></span></p><p><span>函数图象可以将函数的有关性质直观地显现出来，当对数的底数不相同时，可以借助对数函数的图象直观性来理解和寻求解题的思路。</span></p><p><span><br></span></p><p><span><span>例</span>1、<span>若</span>a<span>≠</span>1<span>，</span>b<span>≠</span>1<span>，</span>a<span>＞</span>0<span>，</span>b<span>＞</span>0<span>，且满足关系式</span>log<sub>a</sub>2=<sub><img doc360img-src='http://image109.360doc.com/DownloadImg/2019/06/2608/_2_' data-ratio='0.15054' data-type='gif' data-w='93' _width='-30px' src='http://image109.360doc.com/DownloadImg/2019/06/2608/_2_'></sub><span>，求</span>a<span>，</span>b<span>的值。</span></span></p><p><span><span>分析：已知关系式中的底数不相同，因此可设</span>log<sub>a</sub>2=<sub><img doc360img-src='http://image109.360doc.com/DownloadImg/2019/06/2608/_2_' data-ratio='0.15054' data-type='gif' data-w='93' _width='-30px' src='http://image109.360doc.com/DownloadImg/2019/06/2608/_2_'></sub>=m<span>，转化为指数来来解决</span></span></p><p><span><span>解析：设</span>log<sub>a</sub>2=<sub><img doc360img-src='http://image109.360doc.com/DownloadImg/2019/06/2608/_2_' data-ratio='0.15054' data-type='gif' data-w='93' _width='-30px' src='http://image109.360doc.com/DownloadImg/2019/06/2608/_2_'></sub>=m<span>，则</span></span></p><p><span><sub><img doc360img-src='http://image109.360doc.com/DownloadImg/2019/06/2608/_3_' data-ratio='0.90291' data-type='gif' data-w='103' _width='-30px' src='http://image109.360doc.com/DownloadImg/2019/06/2608/_3_'></sub><span>。</span></span></p><p><span><span>于是有</span>&nbsp;&nbsp;<sub><img doc360img-src='http://image109.360doc.com/DownloadImg/2019/06/2608/_4_' data-ratio='0.24658' data-type='gif' data-w='73' _width='-30px' src='http://image109.360doc.com/DownloadImg/2019/06/2608/_4_'></sub><span>，</span></span></p><p><span><span>因为</span>&nbsp;&nbsp;&nbsp;a<sup>m</sup><span>＞</span>0,</span></p><p><span><span>所以&nbsp;</span><sub><img doc360img-src='http://image109.360doc.com/DownloadImg/2019/06/2608/_5_' data-ratio='0.' data-type='gif' data-w='115' _width='-30px' src='http://image109.360doc.com/DownloadImg/2019/06/2608/_5_'></sub><span>，</span></span></p><p><span><span>于是&nbsp;</span>log<sub>a</sub>2=log<sub>b</sub>3=<span>－１，</span></span></p><p><span><span>解得&nbsp;</span><sub><img doc360img-src='http://image109.360doc.com/DownloadImg/2019/06/2608/_6_' data-ratio='0.52' data-type='gif' data-w='75' _width='-30px' src='http://image109.360doc.com/DownloadImg/2019/06/2608/_6_'></sub><span>。</span></span></p><p><span><span><br></span></span></p><p><span><span>例</span>2、<span>设</span>log<sub>2</sub>3=a<span>，</span>log<sub>3</sub>7=b<span>，求</span>log<sub>42</sub>56<span>的值。</span></span></p><p><span><span>分析：两个已知对数式的底数不相同，无法直接进行计算，所以首先应考虑统一底数，从条件看应该把底数统一为</span>3<span>。</span></span></p><p><span><span>解析：由</span>log<sub>2</sub>3=a<span>，可得</span></span></p><p><span><sub><img doc360img-src='http://image109.360doc.com/DownloadImg/2019/06/2608/_7_' data-ratio='0.6' data-type='gif' data-w='65' _width='-30px' src='http://image109.360doc.com/DownloadImg/2019/06/2608/_7_'></sub><span>，</span></span></p><p><span><span>所以</span><sub><img doc360img-src='http://image109.360doc.com/DownloadImg/2019/06/2608/_8_4292' data-ratio='0.42553' data-type='gif' data-w='235' _width='-30px' src='http://image109.360doc.com/DownloadImg/2019/06/2608/_8_4292'></sub></span></p><p><span><sub><img doc360img-src='http://image109.360doc.com/DownloadImg/2019/06/2608/_9_42934' data-ratio='0.68421' data-type='gif' data-w='76' _width='-30px' src='http://pubimage.360doc.com/wz/default.gif'></sub><span>。</span></span></p><p><span><span><br></span></span></p><p><span><span>例</span>3、<span>若</span>log<sub>a</sub>2<span>＜</span>log<sub>b</sub>2<span>＜</span>0<span>，则</span>a<span>，</span>b<span>满足的关系是（</span>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;<span>）</span></span></p><p><span><span>（</span>A<span>）</span>1<span>＜</span>a<span>＜</span>b</span></p><p><span><span>（</span>B<span>）</span>1<span>＜</span>b<span>＜</span>a</span></p><p><span><span>（</span>C<span>）</span>0<span>＜</span>a<span>＜</span>b<span>＜</span>1</span></p><p><span><span>（</span>D<span>）</span>0<span>＜</span>b<span>＜</span>a<span>＜</span>1</span></p><p><span>分析：两个对数式底数不同，但真数相同，把两个对数式看作是两个对数函数在自变量取同一个值时的两个不同的函数值，可通过图象来分析。</span></p><p><img doc360img-src='http://image109.360doc.com/DownloadImg/2019/06/2608/_10_42965' data-ratio='0.7' data-type='gif' data-w='180' _width='-30px' src='http://pubimage.360doc.com/wz/default.gif'></p><p><span><span>解析：</span>log<sub>a</sub>2<span>，</span>log<sub>b</sub>2<span>可以看成是对数函数</span>y=&nbsp;log<sub>a</sub>x<span>，</span>y=&nbsp;log<sub>b</sub>x<span>在</span>x=2<span>时的两个函数值，可得大致图象（如图）。显然，</span>a<span>，</span>b<span>均小于</span>1<span>，</span></span></p><p><span><span>根据对数函数的底数和图象的关系可得</span>0<span>＜</span>b<span>＜</span>a<span>＜</span>1<span>，故选（</span>D<span>）。</span></span></p><p><span><span><br></span></span></p><p><span><span>二、真数是和差的形式</span></span></p><p><span>利用对数的运算性质可将运算级别较高的运算降底为级别较低的运算，而和与差是运算中的最低级别，所以在处理真数是和差形式的对数问题时，主要有两种处理方法：①整体考虑；②对真数因式分解。</span></p><p><span></span></p><p><span><span>例</span>4、<span>求满足等式</span></span></p><p><span><sub><img doc360img-src='http://image109.360doc.com/DownloadImg/2019/06/2608/_11_' data-ratio='0.' data-type='gif' data-w='264' _width='-30px' src='http://pubimage.360doc.com/wz/default.gif'></sub><span>的</span>x<span>的值。</span></span></p><p><span><span>分析：所给等式出现了对数之和的同时，又出现了一项含有</span>x<span>但又不带对数符号的项，因此直接运用对数的运算法则及相关的性质无法运算，但两个带有对数符号的项的结构相似，因此解答此题要从结构上整体考虑。</span></span></p><p><span><span>解析：由</span><sub><img doc360img-src='http://image109.360doc.com/DownloadImg/2019/06/2608/_11_' data-ratio='0.' data-type='gif' data-w='264' _width='-30px' src='http://pubimage.360doc.com/wz/default.gif'></sub><span>，</span></span></p><p><span><span>得</span><sub><img doc360img-src='http://image109.360doc.com/DownloadImg/2019/06/2608/_12_' data-ratio='0.026481' data-type='gif' data-w='287' _width='-30px' src='http://pubimage.360doc.com/wz/default.gif'></sub><span>，</span></span></p><p><span><span>所以</span>&nbsp;&nbsp;<sub><img doc360img-src='http://image109.360doc.com/DownloadImg/2019/06/2608/_13_' data-ratio='0.' data-type='gif' data-w='116' _width='-30px' src='http://pubimage.360doc.com/wz/default.gif'></sub></span></p><p><span><sub><img doc360img-src='http://image109.360doc.com/DownloadImg/2019/06/2608/_14_' data-ratio='0.' data-type='gif' data-w='167' _width='-30px' src='http://pubimage.360doc.com/wz/default.gif'></sub><span>，</span></span></p><p><span><span>令</span>f(x)=<sub><img doc360img-src='http://image109.360doc.com/DownloadImg/2019/06/2608/_15_' data-ratio='0.' data-type='gif' data-w='145' _width='-30px' src='http://pubimage.360doc.com/wz/default.gif'></sub></span></p><p><span>f(2x)=<sub><img doc360img-src='http://image109.360doc.com/DownloadImg/2019/06/2608/_16_' data-ratio='0.' data-type='gif' data-w='139' _width='-30px' src='http://pubimage.360doc.com/wz/default.gif'></sub><span>，</span></span></p><p><span><span>于是有</span>&nbsp;&nbsp;&nbsp;f(x)=<span>－</span>f(2x)<span>，</span></span></p><p><span><span>易证</span>&nbsp;&nbsp;f(x)<span>是</span>R<span>的减函数，又是奇函数，</span></span></p><p><span><span>故由</span>f(x)=f(<span>－</span>2x)<span>，可得</span></span></p><p><span>x=<span>－</span>2x<span>，</span>x=0<span>。</span></span></p><p><span><span><br></span></span></p><p><span><span>三、对数与对数相乘</span></span></p><p><span>两对数相乘无法利用对数的运算性质求解，因此在解决此类问题时，要根据所给的关系式认真分析其结构特点，主要有三种处理方法：①利用换底公式；②整体考虑；③化各对数为和差的形式。</span></p><p><span><span><br></span></span></p><p><span><span>例</span>5、<span>设</span>log<sub>2</sub>3<span>·</span>log<sub>3</sub>4<span>·</span>log<sub>4</sub>5<span>·</span>log<sub>5</sub>6<span>·</span>log<sub>6</sub>7<span>·</span>log<sub>7</sub>8<span>·</span>log<sub>8</sub>m=log<sub>3</sub>27<span>，求</span>m<span>的值。</span></span></p><p><span><span>分析：已知等式是七个对数之积，其特点是：从第二个对数开始的每一个对数的底数是前一个对数的真数，真数是后一个对数的底数，因此采用换底公式将各对数换成以</span>2<span>为底的两个对数的商，然后约分可达到目的。</span></span></p><p><span>解析：由已知条件得</span></p><p><span>log<sub>2</sub>3<span>·</span>log<sub>3</sub>4<span>·</span>log<sub>4</sub>5<span>·</span>log<sub>5</sub>6<span>·</span>log<sub>6</sub>7<span>·</span>log<sub>7</sub>8<span>·</span>log<sub>8</sub>m</span></p><p><span>=log<sub>2</sub>3<span>·</span><sub><img doc360img-src='http://image109.360doc.com/DownloadImg/2019/06/2608/_17_' data-ratio='0.' data-type='gif' data-w='304' _width='-30px' src='http://pubimage.360doc.com/wz/default.gif'></sub></span></p><p><span>=log<sub>2</sub>m=log<sub>3</sub>27=3</span></p><p><span><span>所以</span>m=8<span>。</span></span></p><p><span><span><br></span></span></p><p><span><span>例</span>6、<span>计算：（</span>lg2<span>）</span><sup>2</sup>lg250+(lg5)<sup>2</sup>lg40<span>。</span></span></p><p><span><span>分析：对数的乘积，无法直接运用对数性质，可以将对数</span>lg250<span>，</span>lg40<span>的真数分解为积的形式，进而将对数转化为和差的形式。</span></span></p><p><span>解析：原式</span></p><p><span>=<span>（</span>lg2<span>）</span><sup>2</sup>lg(5<sup>2</sup><span>×</span>10)+<span>（</span>lg5<span>）</span><sup>2</sup>lg(2<sup>2</sup><span>×</span>10)</span></p><p><span>=(lg2)<sup>2</sup>(2lg5+1)+(lg5)<sup>2</sup>(2lg2+1)</span></p><p><span>=(lg2)<sup>2</sup>+2lg2(lg5)<sup>2</sup>+2lg5(lg2)<sup>2</sup>+(lg5)<sup>2</sup></span></p><p><span>=(lg2)<sup>2</sup>+2lg2lg5(lg5+lg2)+(lg5)<sup>2</sup></span></p><p><span>=(lg2)<sup>2</sup>+2lg2lg5+(lg5)<sup>2</sup></span></p><p><span>=(lg2+lg5)<sup>2</sup>=(lg10)<sup>2</sup>=1<span>。</span></span></p> 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