Решите уравнение log_2(25^х -1) = 2+log_2(5^(х+3)+1)

[latex]log_2(25^{x}-1)=2+log_2(5^{x+3}+1)\; ;\\\\ ODZ:\; \left \{ {{25^{x}-1\ \textgreater \ 0} \atop {5^{x+3}+1\ \textgreater \ 0}} \right. \; \left \{ {{5^{2x}\ \textgreater \ 1} \atop {5^{x}\cdot 5^3\ \textgreater \ -1}} \right. \; \; \Rightarrow \; \; 2x\ \textgreater \ 0\; ,\; x\ \textgreater \ 0\\\\log_2(5^{2x}-1)=log_24+log_2(5^{x+3}+1)\\\\5^{2x}-1=4(5^{x}\cdot 5^3+1)\\\\5^{2x}-500\cdot 5^{x}-5=0\\\\t=5^{x}\ \textgreater \ 0,\; \; \; t^2-500t-5=0\\\\D/4=62505=9\cdot 6945\\\\t_1=250-3\sqrt{6945}\ \textless \ 0\\\\t_2=250+3\sqrt{6945}\ \textgreater \ 0\\\\5^{x}=250+3\sqrt{6945}\\\\x=log_5(250+3\sqrt{6945})[/latex]