ГДЗ по алгебре и начала математического анализа 11 класс Колягин Задание 760

\[\log_{12}3 = a:\]

\[\log_{\sqrt{3}}8 = \log_{\sqrt{3}}2^{3} = 3\log_{\sqrt{3}}2 =\]

\[= 3 \bullet \frac{\log_{2}2}{\log_{2}\sqrt{3}} = \frac{3}{\log_{2}3^{0,5}} =\]

\[= \frac{3}{0,5\log_{2}3} = \frac{6}{\log_{2}3} =\]

\[= 6\ :\frac{\log_{12}3}{\log_{12}2} = 6 \bullet \frac{\log_{12}2}{\log_{12}3} =\]

\[= \frac{6 \bullet \log_{12}4^{0,5}}{\log_{12}3} = \frac{6 \bullet 0,5 \bullet \log_{12}4}{\log_{12}3} =\]

\[= \frac{3 \bullet \log_{12}\frac{12}{3}}{\log_{12}3} =\]

\[= \frac{3\left( \log_{12}12 - \log_{12}3 \right)}{\log_{12}3} =\]

\[= \frac{3(1 - a)}{a}.\]

\[Ответ:\ \ \frac{3(1 - a)}{a}.\]