логарифмы с решением пожалуйста: log 16 по основанию 12 + log 9 по основанию 12= log 363...

$\log_{12} 16 + \log_{12} 9 = \log_{12} 4^2 + \log_{12} 3^2 = 2\log_{12} 4 + 2\log_{12} 3 = 2(\log_{12} 4 + \log_{12} 3) = 2(\log_{12} ({4 * 3})) = 2(\log_{12} 12) = 2$

$\log_{11} 363 - \log_{11} 3 = \log_{11} {11^2*3} - \log_{11} 3 = \log_{11} {11^2} + \log_{11} 3 - \log_{11} 3 = 2\log_{11} {11} = 2$

$7^{\log_7 24} = 24$

$\frac{\log_5 64}{\log_5 4} = \log_4 64 = \log_4 4^2 = 2 \log_4 4 = 2$

$16^{\log_4{(5-\sqrt{5})}} + 4^{\log_2{(5+\sqrt{5})}} = 2^{4{\log_{2^2}{(5-\sqrt{5})}}} + 2^{2\log_2{(5+\sqrt{5})}} = 2^{2{\log_{2}{(5-\sqrt{5})}}} + 2^{2\log_2{(5+\sqrt{5})}} = (2^{{\log_{2}{(5-\sqrt{5})}}})^2 + (2^{\log_2{(5+\sqrt{5})}})^2 = (5-\sqrt{5})^2 + (5+\sqrt{5})^2= 25 - 10\sqrt 5 + 5 + 25 + 10\sqrt 5 + 5 = 60$