Ecuatii exponentiale cu modul

Elena09 Mier Iun 06, 2012 8:40 pm

Sa se rezolve ecuatiile:
$a)2^{|x+1|}=|2^{x}-1|+2^x+1$
$b)6^x=2^x+4^x+2\sqrt{8^{x}}$

**Admin** Mier Iun 06, 2012 10:54 pm

Problema 2

$\begin{array}{l} {6^x} = {2^x} + {4^x} + 2\sqrt {{8^x}} \Leftrightarrow {\left( {{2^{\frac{x}{2}}} + 1} \right)^2} = {3^x} \Leftrightarrow \left( {{2^{\frac{x}{2}}} + 1 - {3^{\frac{x}{2}}}} \right)\left( {{2^{\frac{x}{2}}} + 1 + {3^{\frac{x}{2}}}} \right) = 0\\ \Rightarrow \left\{ \begin{array}{l} {2^{\frac{x}{2}}} - {3^{\frac{x}{2}}} = - 1\left( {{\text{posibil in R}}} \right)\\ {2^{\frac{x}{2}}} + {3^{\frac{x}{2}}} = - 1\left( {{\text{imposibil in R}}} \right) \end{array} \right.\\ {2^{\frac{x}{2}}} - {3^{\frac{x}{2}}} = - 1 \Rightarrow \left\{ \begin{array}{l} f = {2^{\frac{x}{2}}}\\ g = {3^{\frac{x}{2}}} \end{array} \right. \Rightarrow \left( { \uparrow {\text{,Bijective}}} \right) \Rightarrow f - g = - 1 \Leftrightarrow \exists {x_0} = {\text{unic}},\,\,{\text{a}}{\text{.i}}:f\left( {{x_0}} \right) - g\left( {{x_0}} \right) = - 1 \end{array}$
Prin încercări aflam ca x0=2 soluție reala unica !

Voi schimba titlul dvs astfel încât sa fie relevant!

Ecuatii exponentiale cu modul

Ecuatii exponentiale cu modul

Re: Ecuatii exponentiale cu modul