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Логаритми 3

Задаци

Текст задатака објашњених у видео лекцији.

Решити следећи задатак.

пр.4)   ${7^{{{\log }_7}3}} = $

пр.5)   ${10^{0,5 - \log \left( {0,375\sqrt {10} } \right)}} = $

пр.6)   $\left( {{{81}^{\frac{1}{4} - \frac{1}{4}{{\log }_9}4}} + {{25}^{{{\log }_{125}}8}}} \right) \cdot {49^{{{\log }_7}2}} = $

Пр.4

\[{7^{{{\log }_7}3}} = 3\]

Пр.5

\[\begin{gathered}
\boxed{{a^{{{\log }_a}b}} = b} \hfill \\
{10^{0,5 - \log \left( {0,375\sqrt {10} } \right)}} = \frac{{{{10}^{0,5}}}}{{{{10}^{\log \left( {0,375\sqrt {10} } \right)}}}} = \frac{{{{10}^{\frac{1}{2}}}}}{{{{10}^{{{\log }_{10}}\left( {\frac{3}{8}\sqrt {10} } \right)}}}} = \frac{{\sqrt {10} }}{{\frac{3}{8}\sqrt {10} }} = \frac{8}{3} \hfill \\
\end{gathered} \]

Пр.6

\[\begin{gathered}
\left( {{{81}^{\frac{1}{4} - \frac{1}{4}{{\log }_9}4}} + {{25}^{{{\log }_{125}}8}}} \right) \cdot {49^{{{\log }_7}2}} = \hfill \\
= \left( {\frac{{{{81}^{\frac{1}{4}}}}}{{{{81}^{\frac{1}{4}{{\log }_9}4}}}} + {{\left( {{5^2}} \right)}^{{{\log }_{{5^3}}}{2^3}}}} \right) \cdot {\left( {{7^2}} \right)^{{{\log }_7}2}} = \hfill \\
= \left( {\frac{{\sqrt[4]{{81}}}}{{{{\left( {{9^2}} \right)}^{\frac{1}{4}{{\log }_9}4}}}} + {5^{2{{\log }_{{5^3}}}{2^3}}}} \right) \cdot {7^{2{{\log }_7}2}} = \hfill \\
= \left( {\frac{3}{{{9^{\frac{1}{2}{{\log }_9}4}}}} + {5^{2 \cdot \frac{1}{3} \cdot 3{{\log }_5}2}}} \right) \cdot {7^{{{\log }_7}{2^2}}} = \left( {\frac{3}{{{9^{{{\log }_9}\sqrt 4 }}}} + {5^{2{{\log }_5}2}}} \right) \cdot {7^{{{\log }_7}4}} = \hfill \\
= \left( {\frac{3}{2} + {5^{{{\log }_5}{2^2}}}} \right) \cdot 4 = \left( {\frac{3}{2} + 4} \right) \cdot 4 = \frac{{11}}{2} \cdot 4 = 22 \hfill \\
\end{gathered} \]

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