Odpowiedź:
[tex]a)\ \ log_{12}2+log_{12}8+log_{12}9=log_{12}(2\cdot8\cdot9)=log_{12}144=log_{12}12^2=2log_{12}12=\\\\=2\cdot1=2\\\\\\b)\ \ log_{3}\frac{1}{12}+log_{3}\frac{14}{15}+log_{3}\frac{10}{21}=log_{3}(\frac{1}{\not12_{6}}\cdot\frac{\not14^7}{\not15_{3}}\cdot\frac{\not10^2}{21})=log_{3}\frac{\not7^1}{\not18_{9}}\cdot\frac{\not2^1}{\not21_{3}}=log_{3}\frac{1}{27}=\\\\=log_{3}(\frac{1}{3})^3=log_{3}3^{-3}=-3log_{3}3=-3\cdot1=-3[/tex]
[tex]c)\ \ log0,12-log0,3+log25=log0,12+log25-log0,3=log(\frac{\not0,12^{0,4}\cdot25}{\not0,3_{1}})=\\\\=log(0,4\cdot25)=log10=1\\\\\\d)\ \ log_{0,2}0,3-log_{0,2}0,5-log_{0,2}15=log_{0,2}0,3-(log_{0,2}0,5+log_{0,2}15)=\\\\=log_{0,2}(\frac{0,3}{0,5\cdot15})=log_{0,2}(\frac{0,3}{7,5})=log_{0,2}0,04=log_{0,2}0,2^2=2log_{0,2}0,2=2\cdot1=2[/tex]
