Задача 54761 (...

5f6cf18a347ae046112d5363

21.10.2020 12:08:30

( log_(2)3+log_(3)16+4)·(log_(2)3–2·log_(12)3)·log_(3)2–log_(2)3=

5f3ea7e3faf909182968ddd9

21.10.2020 15:18:04

[m]( log_{2}3+log_{3}16+4)\cdot (log_{2}3–2\cdot log_{12}3)\cdot log_{3}2–log_{2}3=[/m]

переходим к одинаковым основаниям у логарифмов

[m]=( \frac {1}{log_{3}2}+log_{3}2^4+4)\cdot(\frac {1}{log_{3}2}–2\cdot \frac{1}{log_{3}12})\cdot log_{3}2– \frac {1}{log_{3}2}=[/m]


[m] =\frac {(2log_{3}2+1)^2}{log_{3}2}\cdot(\frac {1}{log_{3}2}–2\cdot \frac{1}{log_{3}3+log_{3}4})\cdot log_{3}2– \frac {1}{log_{3}2}=[/m]


[m] =(2log_{3}2+1)^2\cdot(\frac {1}{log_{3}2}–\frac{2}{1+2log_{3}2})– \frac {1}{log_{3}2}=[/m]

[m]= (2log_{3}2+1)^2\cdot\frac {1+2log_{3}2-2log_{3}2}{log_{3}2(1+2log_{3}2)}– \frac {1}{log_{3}2}=[/m]

[m]= (2log_{3}2+1)^2\cdot\frac {1}{log_{3}2(1+2log_{3}2)}– \frac {1}{log_{3}2}=[/m]

[m]=\frac {2log_{3}2+1}{log_{3}2}– \frac {1}{log_{3}2}=\frac {2log_{3}2+1-1}{log_{3}2}=log_{3}2[/m]