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Логаритми 1

Логаритми. Дефиниција, особине. Једноставни примери.

Задаци

Текст задатака објашњених у видео лекцији.

Решити логаритам.

пр.1)   ${\log _2}1 = $

           ${\log _7}7 = $

           ${\log _8}2 + {\log _8}4 = $

           ${\log _4}28 - {\log _4}7 = $

           ${\log _3}81 = $

           ${\log _{64}}4 = $

           ${\log _{\sqrt 5 }}\frac{1}{{125}} = $

           ${\log _4}7 \cdot {\log _7}4 = $

           ${\log _2}3 \cdot {\log _3}4 \cdot {\log _4}5 \cdot {\log _5}6 = $

Пр. 1

\[\begin{gathered}
{\log _2}1 = 0 \hfill \\
{\log _7}7 = 1 \hfill \\
{\log _8}2 + {\log _8}4 = {\log _8}2 \cdot 4 = {\log _8}8 = 1 \hfill \\
{\log _4}28 - {\log _4}7 = {\log _4}28 \div 7 = {\log _4}4 = 1 \hfill \\
{\log _3}81 = {\log _3}{3^4} = 4{\log _3}3 = 4 \hfill \\
{\log _{64}}4 = {\log _{{4^3}}}4 = \frac{1}{3}{\log _4}4 = \frac{1}{3} \hfill \\
{\log _{\sqrt 5 }}\frac{1}{{125}} = {\log _{{5^{\frac{1}{2}}}}}{5^{ - 3}} = - 3 \cdot 2{\log _5}5 = - 6 \hfill \\
{\log _4}7 \cdot {\log _7}4 = {\log _4}7\frac{1}{{{{\log }_4}7}} = 1 \hfill \\
{\log _2}3 \cdot {\log _3}4 \cdot {\log _4}5 \cdot {\log _5}6 = \hfill \\
= \frac{{{{\log }_2}3}}{{{{\log }_2}2}} \cdot \frac{{{{\log }_2}4}}{{{{\log }_2}3}} \cdot \frac{{{{\log }_2}5}}{{{{\log }_2}4}} \cdot \frac{{{{\log }_2}6}}{{{{\log }_2}5}} = {\log _2}6 = \hfill \\
= {\log _2}2 \cdot 3 = {\log _2}2 + {\log _2}3 = 1 + {\log _2}3 \hfill \\
\end{gathered} \]