Numere complexe, Regulile de calcul cu logaritmi

Elena09 Mar Iun 12, 2012 1:08 am

Sa se rezolve in C(numere complexe):

a)4z^2 + 8|z|^2=3.

Folosind regulile de calcul cu logaritmi sa se aduca la o forma mai simpla:

a) $log_{2}\sqrt{2} + 6log_{4}\sqrt[6]{2+\sqrt{2}} + \frac{5}{2}log_{2}\sqrt[6]{2-\sqrt{2}}$

b) $3^{1-log_{3}^{10}} - 4^{1-log_{4}^{5}} - 6^{log_{6}^{9}-1}$

**Admin** Mar Iun 12, 2012 2:32 am

Problema 1

$\begin{array}{*{20}{l}} {E = {{\log }_2}\sqrt 2 + 6{{\log }_4}\sqrt[6]{{2 + \sqrt 2 }} + \frac{5}{2}{{\log }_2}\sqrt[6]{{2 - \sqrt 2 }}}\\ {{{\log }_2}\sqrt 2 = {{\log }_2}{2^{\frac{1}{2}}} = \frac{1}{2}}\\ {6{{\log }_4}\sqrt[6]{{2 + \sqrt 2 }} = {{\log }_4}{{\left( {\sqrt[6]{{2 + \sqrt 2 }}} \right)}^6} = {{\log }_4}\left( {2 + \sqrt 2 } \right) = \frac{{{{\log }_2}\left( {2 + \sqrt 2 } \right)}}{{{{\log }_2}4}} = \frac{1}{2}{{\log }_2}\left( {2 + \sqrt 2 } \right)}\\ {\frac{5}{2}{{\log }_2}\sqrt[6]{{2 - \sqrt 2 }} = \frac{6}{2}{{\log }_2}\sqrt[6]{{2 - \sqrt 2 }} - \frac{1}{2}{{\log }_2}\sqrt[6]{{2 - \sqrt 2 }} = \frac{1}{2}\left( {{{\log }_2}\left( {2 - \sqrt 2 } \right) - {{\log }_2}\sqrt[6]{{2 - \sqrt 2 }}} \right)} \end{array}$
$E = \frac{1}{2} + \frac{1}{2}\left( {\underbrace {lo{g_2}\left( {2 - \sqrt 2 } \right) + lo{g_2}\left( {2 + \sqrt 2 } \right)}_{ = 1} - lo{g_2}\sqrt[6]{{2 - \sqrt 2 }}} \right) = 1 - \frac{1}{2}lo{g_2}\sqrt[6]{{2 - \sqrt 2 }} = 1 - lo{g_2}\sqrt[{12}]{{2 - \sqrt 2 }}$
Cam atât se poate face !

**Admin** Mar Iun 12, 2012 2:39 am

Problema b

Daca problema este interpretata corect, atunci o soluție ar fi :
$\begin{array}{l} {3^{1 - log_3^{10}}} - {4^{1 - log_4^5}} - {6^{log_6^9 - 1}}\\ {3^{1 - log_3^{10}}} = \frac{{{3^1}}}{{{3^{log_3^{10}}}}} = \frac{3}{{{{10}^{lo{g_3}3}}}} = \frac{3}{{10}} \end{array}$
Idem restul !

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