Ecuatie logaritmica

lflorea Mier Dec 12, 2012 1:25 am

Rezolvati ecuatia $x^{\log _{}ab\ }=x^{2}-a^{2}+b$ unde a < 1< b.

**Greatmath** Sam Dec 22, 2012 8:09 am

Am văzut deja ca acest subiect a creat valva online .... voi prezenta soluția mea:

$\begin{array}{l} b > 1 > a \Rightarrow \left\{ \begin{array}{l} b > {a^2}\\ {\log _a}b < 0 \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} b - {a^2} > 0\\ 0 \le {x^{{{\log }_a}b}} < 1 \end{array} \right.\\ \\ {x^2} \ge 0 \Rightarrow {x^2} + p = 0 \to {\text{nu are solutii in R daca p}} > {\rm{0}}\\ \\ f\left( x \right) = {x^2} - {a^2} + b > 0,\,\forall x \in R\\ \\ 0 \le g\left( x \right) = {x^{\log \left( {a,b} \right)}} < 1\\ \\ {\rm{In}}\,\,\boxed{\left[ {{\color{Red} 1,\infty} } \right)} \Rightarrow {\text{Nu avem solutii}} \end{array}$

$\begin{array}{l} \boxed{{\color{Red} x = 0}} \Rightarrow 0 = b - {a^2} \Leftrightarrow b = {a^2} \to {\text{imposibila solutie!}}\\ \\ {\text{Deci vom cauta solutii in }}\left( {0,1} \right)\\ \\ \log \left( {a,b} \right) = {\log _x}\left( {{x^2} - {a^2} + b} \right)\\ \\ a = \frac{m}{n},\,m < n \Rightarrow \log \left( {a,b} \right) = \log \left( {\frac{n}{m},b} \right)\,\mathop = \limits^{{\rm{ecuatie}}} \,{\log _x}\frac{1}{{{x^2} - {a^2} + b}},\,x > 1\\ \\ n\left( x \right) = \log \left( {\frac{n}{m},b} \right) = {\rm{monoton - crescator}}{\rm{,}}\,\,{\rm{m}}\left( x \right) = {\log _x}\frac{1}{{{x^2} - {a^2} + b}},\,x > 1 = {\rm{monoton - descrescator}}\\ \\ n\left( x \right) = m\left( x \right) \Leftrightarrow x = {\rm{unic}}.\\ \\ {\text{Se poate observa clar }}\boxed{{\color{Red} \mathbf{x = a}}}{\text{ solutie}\ \mathrm{unica}} \end{array}$

Ecuatie logaritmica

Ecuatie logaritmica

Re: Ecuatie logaritmica