Question

hey! am rezolvat exercitiul dar nu stiu daca este corect rezolvat.
1.Sa se rezolve in multimea numerelor reale,functia: 3^x-3^-x=10/3.
Va multumesc!

Answer 1

$3^{\big{\text{x}}}-3^{-\big{\text{x}}}=\dfrac{10}{3} \\ \\ 3^{\Big{\text{x}}}-\dfrac{1}{3^{\big{\text{x}}}} = \dfrac{10}{3} \\ \\ Notam 3^{\big{\text{x}}} = t, \quad t\ \textgreater \ 0 \\ \\ t-\dfrac{1}{t}=\dfrac{10}{3} \Big|\cdot 3t \\ \\ 3t^2-3 = 10t \\ \\ 3t^2-10t-3 = 0 \\ \Delta = 100-4\cdot 3\cdot (-3) = 100+36 = 136=(2\sqrt{34})^2\\ \\ t_{1,2}= \dfrac{10\pm 2\sqrt{34}}{2\cdot 3} =\dfrac{ 5\pm \sqrt{34}}{3}$

[tex]\bullet$ $t_1 = \dfrac{ 5- \sqrt{34}}{3}\ \textless \ 0 \\ \\ \bullet $ $ t_2 = \dfrac{ 5+\sqrt{34}}{3}\ \textgreater \ 0 \\ \\ \Rightarrow t = \dfrac{ 5+\sqrt{34}}{3} \Rightarrow 3^{\big{\text{x}}} = \dfrac{ 5+\sqrt{34}}{3} \Rightarrow \text{x} = \log_{\big3}\Big(\dfrac{ 5+\sqrt{34}}{3}\Big) \Rightarrow \\ \\ \Rightarrow \text{x} = \log_{\big3}\big( 5+\sqrt{34}}\big) - \log_{\big3}3 \Rightarrow \boxed{x =\log_{\big3}\big( 5+\sqrt{34}\big) -1 }}[/tex]