Помогите, пожалуйста, решить! [tex]log_{6}30/log_{30}6-log_{6}180/log_{5}6[/tex]. Created by Udjin. algebra-ru

Обозначим Тогдаответ: 1

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fuflunce

Обозначим $log_6 5=a$

Тогда

$log_6 30=log_6 (6*5)=log_6 6+log_6 5=a+1;\\ log_{30} 6=\frac{1}{log_ 6 30}=\frac{1}{a+1};\\ log_6 180=log_ 6 (6*30)=log_6 +log_6 30=1+a+1=a+2;\\ log_5 6=\frac{1}{log_6 5}=\frac{1}{a};\\ \frac{log_6 30}{log_{30} 6}-\frac{log_6 180}{log_5 6}=\\ \frac{a+1}{\frac{1}{a+1}}-\frac{a+2}{\frac{1}{a}}=\\ (a+1)^2-a(a+2)=\\ a^2+2a+1-a^2-2a=\\ 1$

ответ: 1

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dtnth

Verified answer

$\frac{log_630}{log_{30}6}-\frac{log_6180}{log_{5}6}=\\ =\frac{log_630}{\frac{1}{log_{6}30}}-\frac{log_6180}{\frac{1}{log_{6}5}}=\\ =log^2_630-log_65*log_6180=\\ = log^2_6(5*6)-log_65*log_6(5*6^2)=\\ =(log_65+log_66)^2-log_65(log_65+log_66^2)=\\ =(log_65+1)^2-log_65(log_65+2)=\\ =log^2_65+2log_65+1-log^2_65-2log_65=1$

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