Question

Sa se rezolve ecuatiile a)49^x-35^x=25^x

Answer 1

$a)49^x-35^x=25^x\\ 49^x-35^x-25^x=0\\ 7^{2x}-(7\cdot 5)^x-5^{2x}=0|:5^{2x}\neq 0\\ \left(\dfrac{7}{5}\right)^{2x}-\left(\dfrac{7}{5}\right)^x -1=0\\ \left(\dfrac{7}{5}\right)^x \stackrel{not }{=} t>0:\\ t^2-t-1=0\\ \Delta=1+4=5\Rightarrow \sqrt{\Delta}=\sqrt 5\\ t_1=\dfrac{1+\sqrt 5}{2}(\text{observatie:fractia aceasta se numeste "numarul de aur"})\\ t_2=\dfrac{1-\sqrt 5}{2}<0(\text{solutia aceasta nu convine}) \\ \text{Prin urmare:}$

$\left(\dfrac{7}{5}\right)^x=\dfrac{1+\sqrt 5}{2}\Rightarrow \boxed{x=\log_{\frac{7}{5}}\left(\dfrac{1+\sqrt5}{2}\right) }$