Ecuatii cu logaritm

Lauryca Mier Iun 06, 2012 8:36 pm

Sa se rezolve ecuatiile:
$a)log _{x^{2}}16 + log_{2x}64=3$
$b)log_{x+2}x+log_{x}(x+2)=\frac{5}{2}$

**Admin** Mier Iun 06, 2012 11:12 pm

Problema 1

$\begin{array}{l} {\rm{lo}}{{\rm{g}}_{{x^2}}}16 + {\rm{lo}}{{\rm{g}}_{2x}}64 = 3 \Leftrightarrow \frac{1}{2}{\rm{lo}}{{\rm{g}}_x}{2^4} + {\rm{lo}}{{\rm{g}}_{2x}}{2^6} = 3 \Leftrightarrow 2{\rm{lo}}{{\rm{g}}_x}2 + 6 \times {\rm{lo}}{{\rm{g}}_{2x}}2 = 3\\ {\rm{lo}}{{\rm{g}}_{2x}}2 = \frac{{{\rm{lo}}{{\rm{g}}_x}2}}{{{\rm{lo}}{{\rm{g}}_x}\left( {2x} \right)}} = \frac{{{\rm{lo}}{{\rm{g}}_x}2}}{{{\rm{lo}}{{\rm{g}}_x}2 + \underbrace {{\rm{lo}}{{\rm{g}}_x}x}_{ = 1}}} = \frac{y}{{y + 1}},\,y\mathop = \limits^{{\rm{notam}}} {\rm{lo}}{{\rm{g}}_x}2 \Rightarrow \\ 2y + 6\left( {\frac{y}{{y + 1}}} \right) = 3 \Rightarrow ... \Rightarrow y = \left\{ { - 3,\frac{1}{2}} \right\} \Rightarrow ... \Rightarrow x = \left\{ {\frac{1}{{\sqrt[3]{2}}},4} \right\} \end{array}$
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**Admin** Joi Iun 07, 2012 2:42 am

Problema 2

$\begin{array}{l} \underbrace {{\rm{lo}}{{\rm{g}}_{x + 2}}x + {\rm{lo}}{{\rm{g}}_x}\left( {x + 2} \right)}_{ = E} = \frac{5}{2}\\ {\rm{lo}}{{\rm{g}}_{x + 2}}x = \frac{{{\rm{lo}}{{\rm{g}}_x}x}}{{{\rm{lo}}{{\rm{g}}_x}x + 2}} = \frac{1}{{{\rm{lo}}{{\rm{g}}_x}\left( {x + 2} \right)}} = \frac{1}{y} \Rightarrow E = \frac{1}{y} + y = \frac{5}{2} \Rightarrow y = \left\{ {\frac{1}{2},2} \right\} \Rightarrow \\ \left\{ \begin{array}{l} {\rm{lo}}{{\rm{g}}_x}\left( {x + 2} \right) = \frac{1}{2}\\ {\rm{lo}}{{\rm{g}}_x}\left( {x + 2} \right) = 2 \end{array} \right. \Rightarrow \left\{ \begin{array}{l} {x^{\frac{1}{2}}} = x + 2 \Rightarrow x \notin R\\ {x^2} = x + 2 \Rightarrow x = 2 = {\rm{unic}} \end{array} \right. \end{array}$
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