помогите пожалуйста!!!!! 1. (3lg2-lg24):(lg3+lg27) 2.(log(3)2+3log(3)0,25):(log(...

sashamels @sashamels

August 2022 1 9 Report

помогите пожалуйста!!!!!

1. (3lg2-lg24):(lg3+lg27)

2.(log(3)2+3log(3)0,25):(log(3)28-log(3)7)

(log(3)2-т.е логарифм числа 2 по основанию 3)

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dtnth

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$(3lg2-lg 24):(lg 3+lg 27)=\\ (lg2^3-lg 24):(lg (3*27))=\\ (lg 8-lg 24):(lg (81))=\\ (lg \frac{8}{24}):(lg (81))=lg \frac{1}{3}:lg 81=\\ log_{81} \frac{1}{3}=\\ log_{3^4} 3^{-1}=\\ -\frac{1}{4} log_3 3=-\frac{1}{4}*1=-0.25$

$(log_3 2+3log_3 0.25):(log_3 28-log_3 7)=\\ (log_3 2+log_3 0.25^3):(log_3 \frac{28}{7})=\\ (log_3 2+log_3 0.015625):(log_3 4)=\\ log_3 (2*0.015625):(log_3 4)=\\ log_3 (2*0.015625):(log_3 4)=\\ log_3 0.03125 :(log_3 4)=\\ log_4 0.03125=log_{2^2} 2^{-5}=\frac{-5}{2} log_2 2=\ -2.5*1=-2.5$

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