Własności logarytmów Dla $a,x,y>0$  i  $a\ne 1$ zachodzą następujące równości: ${\log }_{a}x+{\log }_{a}y={\log }_a\left(x\cdot y\right)$  ${\log }_{a}x-{\log }_{a}y={\log }_a\frac{x}{y}$

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a)

Obliczamy:

${\log }_{12}2+{\log }_{12}8+{\log }_{12}9={\log }_{12}\left(2\cdot 8\cdot 9\right)={\log }_{12}144$

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b)

Obliczamy:

${\log }_3\frac{1}{12}+{\log }_3\frac{14}{15}+{\log }_3\frac{10}{21}={\log }_3\left(\frac{1}{12}\cdot \frac{{\bcancel{14}}^2}{{\cancel{15}}_3}\cdot \frac{{\cancel{10}}^2}{{\bcancel{21}}_3}\right)=$

$={\log }_3\left(\frac{1}{12}\cdot \frac{2}{3}\cdot \frac{2}{3}\right)={\log }_3\frac{1}{27}=-3$

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c)

Obliczamy:

$\log 0,12-\log 0,3+{\log }_{}25=\log \left(\frac{12}{100}:\frac{3}{10}\right)+\log 25=$

$=\log \left(\frac{12}{100}\cdot \frac{10}{3}\right)+\log 25=\log \frac{4}{10}+\log 25=$

$=\log \left(\frac{2}{5}\cdot 25\right)={\log }_{10}10=1$

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d)

Obliczamy:

${\log }_{0,2}0,3-{\log }_{0,2}0,5-{\log }_{0,2}15={\log }_{0,2}\left(\frac{3}{10}:\frac{5}{10}\right)-{\log }_{0,2}15=$

$={\log }_{0,2}\left(\frac{3}{10}\cdot \frac{10}{5}\right)-{\log }_{0,2}15={\log }_{0,2}\frac{3}{5}-{\log }_{0,2}15=$

$={\log }_{0,2}\left(\frac{3}{5}:15\right)={\log }_{0,2}\left(\frac{3}{5}\cdot \frac{1}{15}\right)={\log }_{\frac{1}{5}}\frac{1}{25}=2$