Логаритми 2

Текст задатака објашњених у видео лекцији.

пр.2)   Логаритмовати израз:

            $A = \frac{{\sqrt {x\sqrt {xy} } }}{{\sqrt[3]{{xy}}}}$

пр.3)   Одредити $x = ?$, ако је

            $\operatorname{logx}  = \frac{1}{2}\left( {\operatorname{loga}  - \operatorname{logb}  + \frac{1}{3}\left( {\operatorname{logc}  - \frac{1}{2}\log d} \right)} \right)$

Пр. 2

\[\begin{gathered}
\log A = \log \frac{{\sqrt {x\sqrt {xy} } }}{{\sqrt[3]{{xy}}}} = \log \sqrt {x\sqrt {xy} } - \log \sqrt[3]{{xy}} = \hfill \\
= \log {\left( {x\sqrt {xy} } \right)^{\frac{1}{2}}} - \log {\left( {xy} \right)^{\frac{1}{3}}} = \frac{1}{2}\log x\sqrt {xy} - \frac{1}{3}\log xy = \hfill \\
= \frac{1}{2}\left( {\log x + \log {{\left( {xy} \right)}^{\frac{1}{2}}}} \right) - \frac{1}{3}\left( {\log x + \log y} \right) = \hfill \\
= \frac{1}{2}\left( {\log x + \frac{1}{2}\log \left( {xy} \right)} \right) - \frac{1}{3}\left( {\log x + \log y} \right) = \hfill \\
= \frac{1}{2}\left( {\log x + \frac{1}{2}\left( {\operatorname{logx} + logy} \right)} \right) - \frac{1}{3}\left( {\log x + \log y} \right) = \hfill \\
= \frac{1}{2}\log x + \frac{1}{4}\operatorname{logx} + \frac{1}{4}logy - \frac{1}{3}\log x - \frac{1}{3}\log y = \hfill \\
= \left( {\frac{1}{2} + \frac{1}{4} - \frac{1}{3}} \right)\log x + \left( {\frac{1}{4} - \frac{1}{3}} \right)\log y = \frac{5}{{12}}\log x - \frac{1}{{12}}\log y \hfill \\
\end{gathered} \]

Пр. 3

\[\begin{gathered}
\operatorname{logx} = \frac{1}{2}\left( {\operatorname{loga} - \operatorname{logb} + \frac{1}{3}\left( {\operatorname{logc} - \frac{1}{2}\log d} \right)} \right) = \hfill \\
= \frac{1}{2}\left( {\operatorname{loga} - \operatorname{logb} + \frac{1}{3}\left( {\operatorname{logc} - \log \sqrt d } \right)} \right) = \hfill \\
= \frac{1}{2}\left( {\operatorname{loga} - \operatorname{logb} + \frac{1}{3}\left( {\log \frac{c}{{\sqrt d }}} \right)} \right) = \hfill \\
= \frac{1}{2}\left( {\operatorname{loga} - \operatorname{logb} + \left( {\log \sqrt[3]{{\frac{c}{{\sqrt d }}}}} \right)} \right) = \hfill \\
= \frac{1}{2}\left( {\log \frac{a}{b} + \left( {\log \sqrt[3]{{\frac{c}{{\sqrt d }}}}} \right)} \right) = \hfill \\
= \frac{1}{2}\left( {\log \frac{a}{b} \cdot \sqrt[3]{{\frac{c}{{\sqrt d }}}}} \right) = \log \sqrt {\frac{a}{b} \cdot \sqrt[3]{{\frac{c}{{\sqrt d }}}}} \hfill \\
\log x = \log \sqrt {\frac{a}{b} \cdot \sqrt[3]{{\frac{c}{{\sqrt d }}}}} \hfill \\
\end{gathered} \]

Ако $\log a = \log b,$ то а=b.

Онда 

\[x = \sqrt {\frac{a}{b} \cdot \sqrt[3]{{\frac{c}{{\sqrt d }}}}} \]