对数例题集
第一题
\[2(\lg_{}{\sqrt{2} })^2+\lg_{}{\sqrt[]{2}} \times \lg_{}{5}
+\sqrt[]{ (\lg_{}{\sqrt[]{2}})^2 - \lg_{}{2} + 1} \]
\[\\ \\
\]
\[\lg_{}{\sqrt[]{2}}[2(\lg_{}{\sqrt[]{2}})+\lg_{}{5}]
+\sqrt[]{(\frac{1}{2}\lg_{}{2})^2-\lg{}{2}+1} \]
\[\\ \\
\]
\[\because (\frac{1}{2}\lg_{}{2})^2-\lg{}{2}+1 符合完全平方公式
\]
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\]
\[\because (\frac{1}{2}\lg_{}{2})^2-\lg{}{2}+1
\Rightarrow (\frac{1}{2}\lg_{}{2}-1) < 0\]
\[\\ \\
\]
\[\because 1-\frac{1}{2} \lg_{}{2} >0
\]
\[\\ \\
\]
\[\therefore (\frac{1}{2}\lg_{}{2})^2-\lg{}{2}+1\Rightarrow 1-\lg_{}{2}+(\frac{1}{2} \lg_{}{2})^2
\]
\[\\ \\
\]
\[\therefore \lg_{}{\sqrt[]{2}}[2(\lg_{}{\sqrt[]{2}})+\lg_{}{5}]
+\sqrt[]{1-\lg{}{2}+(\frac{1}{2}\lg_{}{2})^2}\]
\[\\ \\
\]
\[\lg_{}{\sqrt[]{2}}[2(\lg_{}{\sqrt[]{2}})+\lg_{}{5}]
+\sqrt[]{(1-\frac{1}{2}\lg_{}{2})^2}\]
\[\\ \\
\]
\[\lg_{}{\sqrt[]{2}}(\lg_{}{\sqrt[]{2}}^2+\lg_{}{5})
+\sqrt[]{(1-\frac{1}{2}\lg_{}{2})^2}\]
\[\\ \\
\]
\[\lg_{}{\sqrt[]{2}}(\lg_{}{2}+\lg_{}{5})+1-\frac{1}{2}\lg_{}{2}
\]
\[\\ \\
\]
\[\lg_{}{\sqrt[]{2}}[\lg_{}{(2\cdot5)}]+1-\frac{1}{2}\lg_{}{2}
\]
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\]
\[\lg_{}{\sqrt[]{2}}(\lg_{}{10})+1-\lg_{}{2}^\frac{1}{2}
\]
\[\\ \\
\]
\[\because \lg{10} = 1
\]
\[\\ \\
\]
\[\therefore \lg{\sqrt[]{2}}+1-\lg{\sqrt[]{2}}=1
\]
第二题
\[(\lg{5})^2+\lg{2} \cdot \lg{50}
\]
\[\\ \\
\]
\[(\lg{5})^2+\lg{2} \cdot \lg{(5\cdot10)}
\]
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\]
\[(\lg{5})^2+\lg{2} \cdot \lg{5}+\lg{2}\cdot\lg{10}
\]
\[\\ \\
\]
\[\lg{5}\cdot\lg{5}+\lg{2} \cdot \lg{5}+\lg{2}
\]
\[\\ \\
\]
\[\lg{5}(\lg{5}+\lg{2} + \frac{\lg{2}}{\lg{5}})
\]
\[\\ \\
\]
\[\lg{5}[\lg{(5\cdot2)} + \frac{\lg{2}}{\lg{5}}]
\]
\[\\ \\
\]
\[\lg{5}(1 + \frac{\lg{2}}{\lg{5}})
\]
\[\\ \\
\]
\[\lg{5} + \lg{5} \cdot \frac{\lg{2}}{\lg{5}}
\]
\[\\ \\
\]
\[\lg{5}+\lg{2}= 1
\]
第三题
\[(\lg{2})^{3}+(\lg{5})^{3}+3\lg{2} \cdot \lg{5}
\]
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\]
\[设\lg{2}=a,\lg{5}=b
\]
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\]
\[=a^{3}+b^{3}+3ab
\]
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\]
\[\because a^3+b^3 \Rightarrow (a+b)(a^2-ab+b^2)
\]
\[\\ \\
\]
\[\therefore a^{3}+b^{3}+3ab \Rightarrow (a+b)(a^{2}-ab+b^{2})+3ab
\]
\[\\ \\
\]
\[\because \lg{2}+\lg{5}=\lg{(2\cdot5)}=1
\]
\[\\ \\
\]
\[(a+b)(a^2-ab+b^2+3ab)=(a+b)(a^{2}+2ab+b^{2})=(a+b)(a+b)^{2}
\]
\[\\ \\
\]
\[(\lg{2}+\lg{5})(\lg{2}+\lg{5})^{2}=1
\]
第四题 定义法示例一
\[已知 \log_{a}{(x+y)}=\frac{1}{3}, \log_{a}{x}=4, 求\log_{a}{y}
\\
即已知 a^{ \frac{1}{3} }=x+y, a^{4}= x, 若 a^{n}= y,求n的值 \]
\[\\
\]
\[\because y=(x+y)-x
\]
\[\\
\]
\[\therefore a^{\frac{1}{3}}-a^{4}=y
\]
\[\\
\]
\[\log_{a}{y}=\log_{a}{a^{\frac{1}{3}}}-\log_{a}{a^{4}}=\log_{a}
{(\frac{a^{\frac{1}{3}}} {a^{4}})} \]
\[\\
\]
\[\log_{a}{y}=\log_{a}{(a^{ -\frac{11}{3}})}
\]
\[\\
\]
\[y=a^{-\frac{11}{3}}
\]
\[\\
\]
\[\therefore \log_{a}{y}=-\frac{11}{3}
\]
第五题 定义法示例二
\[若\log_{4}{(3x-1)}=\log_{4}{(x-1)}+\log_{4}{(x+3)},求x \\
解:\\
\because \log_{4}{(3x-1)}=\log_{4}{[(x-1)(x+3)]} \\
\therefore 3x-1=(x-1)(x+3) \Rightarrow 3x-1=x^{2}+2x-3 \\
x^{2}-x-2=0 \\
\]
然后使用十字相乘法求得解,注意真数的定义域大于零,
利用对数的定义完成对数式与指数式的互相转换是定义法的关键
第六题
\[\log_{a}{x}=2,\log_{b}{x}=3,\log_{c}{x}=6,求\log_{abc}{x}\\
\\
设 r=\log_{abc}{x}\]
\[\\ \\
\]
\[已知 \log_{a}{x}=\frac{\lg_{}{x}}{\lg_{}{a}},
\log_{b}{x}=\frac{\lg_{}{x}}{\lg_{}{b}},
\log_{c}{x}=\frac{\lg_{}{x}}{\lg_{}{c}}
\]
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\]
\[\log_{abc}{x}=\frac{\lg_{}{x}}{\lg_{}{abc}}
\]
\[\\ \\
\]
\[而 \lg_{}{abc}=\lg_{}{a}+\lg_{}{b}+\lg_{}{c}
\]
\[\\ \\
\]
\[\therefore r=\frac{\lg_{}{x}}{\lg_{}{a}+\lg_{}{b}+\lg_{}{c}}
\]
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\]
\[\frac{1}{r}= \frac{1}{ \frac{\lg_{}{x}}{\lg_{}{a}+\lg_{}{b}+\lg_{}{c}} }
\]
\[\\ \\
\]
\[\Rightarrow
\frac{\lg_{}{a}}{\lg_{}{x}} + \frac{\lg_{}{b}}{\lg_{}{x}}+ \frac{\lg_{}{c}}{\lg_{}{x}}
\]
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\]
\[再根据对数的倒数关系式:\frac{1}{r}= \frac{1}{2}+\frac{1}{3}+\frac{1}{6} =1
\]
\[\\ \\
\]
\[\therefore \log_{abc}{x}=1
\]
第七题
\[若 \log_{18}{9}=a, 18^{b}=5,如何用a,b表示 \log_{36}{45}
\]
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\]
\[\because \log_{36}{45}=\frac{\log_{18}{45}}{\log_{18}{36}}
\]
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\]
\[\log_{18}{45}=\log_{18}{(5 \cdot 9)} \Rightarrow \log_{18}{9}+\log_{18}{5}
\Rightarrow a+b
\]
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\]
\[\log_{18}{36}=\log_{18}{(\frac{36 \cdot 9}{9})}
\Rightarrow
\log_{18}{(36\cdot9)} - \log_{18}{9}
\]
\[\\ \\
\]
\[\because \log_{18}{(36\cdot9)} \Rightarrow
\log_{18}{(4\cdot9\cdot9)}
\]
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\]
\[设 \log_{18}{(4\cdot9\cdot9)}=n, 则 18^{n}=2^{2} \cdot 9^{2}, n=2
\]
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\]
\[\therefore \log_{36}{45}=\frac{a+b}{2-a}
\]
