$\text{a)}L=\frac{{\log }_{2}50-{\log }_{2}5}{{\log }_{2}45-{\log }_{2}9}=\frac{{\log }_2\left(\frac{50}{5}\right)}{{\log }_2\left(\frac{45}{9}\right)}=\frac{{\log }_{2}10}{{\log }_{2}5}={\log }_{5}10=\frac{\log 10}{\log 5}=\frac{1}{\log 5}=P$  

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$\text{b)}\frac{{\log }_{3}20-{\log }_{9}25}{{\log }_{\sqrt{3}}2+{\log }_{3}2}=\frac{{\log }_{3}20-\frac{{\log }_{3}25}{{\log }_{3}9}}{\frac{{\log }_{3}2}{{\log }_3\sqrt{3}}+{\log }_{3}2}=\frac{{\log }_{3}20-\frac{1}{2}{\log }_{3}25}{\frac{{\log }_{3}2}{\frac{1}{2}}+{\log }_{3}2}=\frac{{\log }_{3}20-{\log }_3{25}^{\frac{1}{2}}}{2{\log }_{3}2+{\log }_{3}2}=$  

$\frac{{\log }_{3}20-{\log }_{3}5}{{\log }_{3}4+{\log }_{3}2}=\frac{{\log }_{3}4}{{\log }_{3}8}={\log }_{8}4=\frac{{\log }_{2}4}{{\log }_{2}8}=\frac{2}{3}$

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$\text{c)}\frac{{\log }_{5}100+{\log }_{2}100}{{\log }_{5}100\cdot {\log }_{2}100}$  

Zamieńmy logarytmy w następujący sposób:

${\log }_{5}100=\frac{\log 100}{\log 5}$  

${\log }_{2}100=\frac{\log 100}{\log 2}$  

 

$=\frac{\frac{\log 100}{\log 5}+\frac{\log 100}{\log 2}}{\frac{\log 100}{\log 5}\cdot \frac{\log 100}{\log 2}}=\frac{\frac{2}{\log 5}+\frac{2}{\log 2}}{\frac{4}{\log 5\cdot \log 2}}=\left(\frac{2\log 2}{\log 5\log 2}+\frac{2\log 5}{\log 5\log 2}\right)\cdot \frac{\log 5\cdot \log 2}{4}=$  

$\frac{\log 4+\log 25}{\log 5\log 2}\cdot \frac{\log 5\log 2}{4}=\frac{\log \left(4\cdot 25\right)}{4}=\frac{\log 100}{4}=\frac{2}{4}=\frac{1}{2}$  

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$\text{d)}\frac{{\log }_{4}12\cdot {\log }_{3}12}{{\log }_{4}12+{\log }_{3}12}$  

Zamieńmy logarytmy:

${\log }_{4}12=\frac{\log 12}{\log 4}$  

${\log }_{3}12=\frac{\log 12}{\log 3}$  

 

$=\frac{\frac{\log 12}{\log 4}\cdot \frac{\log 12}{\log 3}}{\frac{\log 12}{\log 4}+\frac{\log 12}{\log 3}}=\frac{\frac{{\log }^{2}12}{\log 3\log 4}}{\log 12\left(\frac{1}{\log 4}+\frac{1}{\log 3}\right)}=\frac{\frac{{\log }^{2}12}{\log 3\log 4}}{\log 12\cdot \frac{\log 3+\log 4}{\log 3\log 4}}=\frac{\frac{{\log }^{2}12}{\log 3\log 4}}{\log 12\cdot \frac{\log 12}{\log 3\log 4}}=\frac{\frac{{\log }^{2}12}{\log 3\log 4}}{\frac{{\log }^{2}12}{\log 3\log 4}}=1$