[tex]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:}[/tex]
[tex]\left(\dfrac{7}{5}\right)^x=\dfrac{1+\sqrt 5}{2}\Rightarrow \boxed{x=\log_{\frac{7}{5}}\left(\dfrac{1+\sqrt5}{2}\right) }[/tex]
