Question

Rezolvati in R ecuatia:
$25^X - 4 * 5^X - 5 = 0$

Answer 1

$25^{\big{\text{x}}}-4\cdot 5^{\big{\text{x}}}-5 = 0 \\ \\ \Big(5^{\big2}\Big)^{\big{\text{x}}}-4\cdot 5^{\big{\text{x}}}-5 = 0 \\ \\ 5^{^{\big{2\cdot\text{x}}}}-4\cdot 5^{\big{\text{x}}}-5 = 0 \\ \\ \Big(5^{\big{\text{x}}}\Big)^{\big2}-4\cdot 5^{\big{\text{x}}}-5 = 0\\ \\ Notam 5^{\big{\text{x}}} = t,\quad t\ \textgreater \ 0. \\ \\ \left\| \begin{array}{c}t^{\big2}-4t-5 = 0\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \\ \Delta = (-4)^2-4\cdot 1\cdot (-5) = 16+20 = 36\end{array} \right| \Rightarrow$

$\Rightarrow \left\| \begin{array}{c} t_1 = \dfrac{4+\sqrt{36}}{2} \\ \\ t_2 = \dfrac{4-\sqrt{36}}{2} \end{array} \right \Rightarrow \left\| \begin{array}{c} t_1 = \dfrac{4+6}{2} \\ \\ t_2 = \dfrac{4-6}{2} \end{array} \right \Rightarrow\left\| \begin{array}{c} t_1 = 5 \\ \\ t_2 = -1\ \textless \ 0\end{array} \right \Rightarrow \\ \\ \\\Rightarrow t=5\Rightarrow 5^{\big{\text{x}}} = 5 \Rightarrow 5^{\big{\text{x}}} = 5^1 \Rightarrow \text{x} = 1 \Rightarrow \boxed{S = \Big\{1\Big\}}$