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[Гост]

IMG-47f0aa629eb8eea2087f2343b7d22ef4-V.png (95.5 KiB) Прегледано 392 пъти

[mail_dinko]

[tex]2 \sqrt {log_a b}((log _a \sqrt [4] {ab}+log _b \sqrt [4] {ab})^{\frac 12}-(log _a \sqrt [4] {\frac{b}{a}}+log _b \sqrt [4] {\frac{a}{b}})^{\frac 12})=[/tex]
[tex]=2 \sqrt {log_a b}((\frac {1}{log _{\sqrt [4] {ab}} a} +\frac {1}{log _{\sqrt [4] {ab}}b} )^{\frac 12}-(\frac {1}{log _{ \sqrt [4] {\frac{b}{a}}}a}+ \frac {1}{log _{\sqrt [4] {\frac {a}{b}}}b})^{\frac 12})=[/tex]
[tex]=2 \sqrt {log_a b}((\frac {{log _{\sqrt [4] {ab}} a} +log _{\sqrt [4] {ab}}b}{{log _{\sqrt [4] {ab}} a} .log _{\sqrt [4] {ab}}b} )^{\frac 12}-(\frac {1}{log _{ \sqrt [4] {\frac{b}{a}}}a}+ \frac {1}{log _{(\frac {b}{a})^{-\frac14}}b})^{\frac 12})=[/tex]
[tex]=2 \sqrt {log_a b}((\frac {{log _{\sqrt [4] {ab}} (ab)} }{{log _{\sqrt [4] {ab}} a} .log _{\sqrt [4] {ab}}b} )^{\frac 12}-(\frac {1}{log _{ \sqrt [4] {\frac{b}{a}}}a}- \frac {1}{log _{(\frac {b}{a})^{\frac14}}b})^{\frac 12})=[/tex]
[tex]=2 \sqrt {log_a b}((\frac {{log _{\sqrt [4] {ab}} (ab)} }{{log _{\sqrt [4] {ab}} a} .log _{\sqrt [4] {ab}}b} )^{\frac 12}-(\frac {1}{log _{ \sqrt [4] {\frac{b}{a}}}a}- \frac {1}{log _{\sqrt [4]{\frac {b}{a}}}b})^{\frac 12})=[/tex]
[tex]=2 \sqrt {log_a b}((\frac {{log _{\sqrt [4] {ab}} (ab)} }{{log _{\sqrt [4] {ab}} a} .log _{\sqrt [4] {ab}}b} )^{\frac 12}-(\frac {log _{\sqrt [4]{\frac {b}{a}}}b -log _{ \sqrt [4] {\frac{b}{a}}}a }{log _{ \sqrt [4] {\frac{b}{a}}}a.log _{\sqrt [4]{\frac {b}{a}}}b})^{\frac 12})=[/tex]
[tex]=2 \sqrt {log_a b}(\sqrt {\frac {{log _{\sqrt [4] {ab}} (ab)} }{{log _{\sqrt [4] {ab}} a} .log _{\sqrt [4] {ab}}b} }-\sqrt {\frac {log _{\sqrt [4]{\frac {b}{a}}}\frac {b}{a} }{log _{ \sqrt [4] {\frac{b}{a}}}a.log _{\sqrt [4]{\frac {b}{a}}}b}})=[/tex]
[tex]=2 \sqrt {log_a b}(\sqrt {\frac {{log _{ ({ab})^{\frac14}} (ab)^1} }{{log _{\sqrt [4] {ab}} a} .log _{\sqrt [4] {ab}}b} }-\sqrt {\frac {log _{{(\frac {b}{a})^{\frac14}}}(\frac {b}{a})^1 }{log _{ \sqrt [4] {\frac{b}{a}}}a.log _{\sqrt [4]{\frac {b}{a}}}b}})=[/tex]
[tex]=2 \sqrt {log_a b}(\sqrt {\frac {4 }{{log _{\sqrt [4] {ab}} a} .log _{\sqrt [4] {ab}}b} }-\sqrt {\frac {4 }{log _{ \sqrt [4] {\frac{b}{a}}}a.log _{\sqrt [4]{\frac {b}{a}}}b}})=[/tex]
[tex]=4 \sqrt {log_a b} (\frac {1}{\sqrt {{log _{\sqrt [4] {ab}} a} .log _{\sqrt [4] {ab}}b}}-\frac {1}{\sqrt {log _{ \sqrt [4] {\frac{b}{a}}}a.log _{\sqrt [4]{\frac {b}{a}}}b}})[/tex]
Аз съм до тук
Надолу не съм сигурен, че е възможно, не видях такова св-во на лог
[tex]4 \sqrt {log_a b} (\frac {1}{\sqrt {log _{\sqrt [4] {ab}}(ab)}}-\frac {1}{\sqrt {log _{\sqrt [4]{\frac {b}{a}}}(ab)}})=[/tex]
[tex]=4 \sqrt {log_a b} (\frac {1}{2}-\frac {1}{\sqrt {log _{\sqrt [4]{\frac {b}{a}}}(ab)}})=[/tex]

[ammornil]

[tex]\log_{a}{\sqrt[4]{ab}}=\log_{a}(a.b)^{\frac{1}{4}}=\log_{a}{a^{\frac{1}{4}}}+\log_{a}{b^{\frac{1}{4}}}=\frac{1}{4}\log_{a}{a}+\frac{1}{4}\log_{a}{b}=\frac{1}{4}(1+\log_{a}{b})[/tex]
[tex]\log_{b}{\sqrt[4]{ab}}=\log_{b}(a.b)^{\frac{1}{4}}=\log_{b}{a^{\frac{1}{4}}}+\log_{b}{b^{\frac{1}{4}}}=\frac{1}{4}\log_{b}{a}+\frac{1}{4}\log_{b}{b}=\frac{1}{4}(1+\log_{b}{a})=\frac{1}{4}\left( 1+\frac{1}{\log_{a}{b}} \right)[/tex]

$$ (1) \hspace{5em} (\log_{a}{\sqrt[4]{ab}}+\log_{b}{\sqrt[4]{ab}})^{\frac{1}{2}}=\frac{1}{2}\left(1+\log_{a}{b}+1+\frac{1}{\log_{a}{b}} \right)^{\frac{1}{2}}=\frac{1}{2}\sqrt{2+\log_{a}{b}+\frac{1}{\log_{a}{b}}} $$

[tex]\log_{a}{\sqrt[4]{\frac{b}{a}}}=\log_{a}{\frac{b^{\frac{1}{4}}}{a^{\frac{1}{4}}}}=\frac{1}{4}\log_{a}{b}-\frac{1}{4}\log_{a}{a}=\frac{1}{4}(\log_{a}{b}-1)[/tex]
[tex]\log_{b}{\sqrt[4]{\frac{b}{a}}}=\log_{b}{\frac{b^{\frac{1}{4}}}{a^{\frac{1}{4}}}}=\frac{1}{4}\log_{b}{b}-\frac{1}{4}\log_{b}{a}=\frac{1}{4}(1-\log_{b}{a})[/tex]

$$ (2) \hspace{8em} \left(\log_{a}{\sqrt[4]{\frac{b}{a}}}+\log_{b}{\sqrt[4]{\frac{b}{a}}}\right)^{\frac{1}{2}}=\frac{1}{2}\left(\log_{a}{b}-1+1-\log_{b}{a} \right)^{\frac{1}{2}}=\frac{1}{2}\sqrt{\log_{a}{b}-\frac{1}{\log_{a}{b}}} $$


$$ 2\sqrt{\log_{a}{b}}\left( \frac{1}{2}\sqrt{2+\log_{a}{b}+\frac{1}{\log_{a}{b}}}- \frac{1}{2}\sqrt{\log_{a}{b}-\frac{1}{\log_{a}{b}}} \right)=\sqrt{\log^{2}_{a}{b}+2\log_{a}{b}+1}-\sqrt{\log^{2}_{a}{b}-1}=|(\log_{a}{b}+1)|-\sqrt{(\log_{a}{b}-1)(\log_{a}{b}+1)} $$

[S.B.]

[tex]2 \sqrt{ \log_{a } b} [ ( \log_{a } \sqrt[4]{ab} + \log_{b } \sqrt[4]{ab}) ^{ \frac{1}{2} } - ( \log_{a } \sqrt[4]{ \frac{b}{a} } + \log_{b } \sqrt[4]{ \frac{a}{b} }) ^{ \frac{1}{2} }] =[/tex]

[tex]= 2 \sqrt{ \log_{a }b }[ ( \log_{a } (ab)^{ \frac{1}{4} } + \log_{b } ab^{ \frac{1}{4} }) ^{ \frac{1}{2} } - ( \log_{a } ( \frac{b}{a}) ^{ \frac{1}{4} } + \log_{b } ( \frac{a}{b}) ^{ \frac{1}{4} }) ^{ \frac{1}{2} }] =[/tex]

[tex]= 2 \sqrt{ \log_{a }b } [ ( \frac{1}{4} \log_{a }a + \frac{1}{4} \log_{a }b + \frac{1}{4} \log_{b }a+ \frac{1}{4} \log_{b }b) ^{ \frac{1}{2} } - ( \frac{1}{4} \log_{a }b - \frac{1}{4} \log_{a }a + \frac{1}{4} \log_{b }a - \frac{1}{4} \log_{b }b )^{ \frac{1}{2} }] =[/tex]

[tex]= 2 \sqrt{ \log_{a }b }[ ( \frac{1}{4} ( \log_{a }b + log_{b }a) + \frac{1}{2}) ^{ \frac{1}{2} } - ( \frac{1}{4}( \log_{a }b + \log_{b }a) - \frac{1}{2}) ^{ \frac{1}{2} } ][/tex]

Преминавам към основа $a$ и получавам:
$$\log_{b }a = \frac{ log_{a } a}{ \log_{a } b} \Rightarrow \log_{b }a = \frac{1}{ \log_{a } b} $$

[tex]2 \sqrt{ \log_{a }b }[ ( \frac{1}{4}( \log_{a }b + \frac{1}{ \log_{a }b } )+ \frac{1}{2}) ^{ \frac{1}{2} } - ( \frac{1}{4}( \log_{a }b + \frac{1}{ \log_{a }b } )- \frac{1}{2}) ^{ \frac{1}{2} }][/tex]

Полагам:[tex]\log_{a } b = t[/tex] и след заместване получавам:

[tex]2 \sqrt{t}[ ( \frac{1}{4}(t + \frac{1}{t}) + \frac{1}{2}) ^{ \frac{1}{2} } - ( \frac{1}{4}( t + \frac{1}{t}) - \frac{1}{2}) ^{ \frac{1}{2} }] =[/tex]

[tex]= 2 \sqrt{t} (\sqrt{ \frac{ t^{2 } + 2t + 1 }{4t} } - \sqrt{ \frac{ t^{2 } - 2t + 1 }{4t} } ) = 2 \sqrt{t} ( \sqrt{ \frac{ (t + 1)^{2 } }{4t} } - \sqrt{ \frac{ (t - 1)^{2 } }{4t} }) =[/tex]

[tex]= 2 \sqrt{t}( \frac{t + 1 - t + 1}{2 \sqrt{t} }) =[/tex]

$= 2$