Решить 49^x-5*7^x+6=0.

[latex]49^x-5\cdot7^x+6 = 0\\ 7^{2x}- 5\cdot7^x + 6 =0\\ y = 7^x\\ y^2-5y+6 = 0\\ (y-2)(y-3) = 0\\ \left [ {{y=2} \atop {y=3}} \right. \\ \left [ {{7^x=2} \atop {7^x=3}} \right. \\ \left [ {{x=\log_72} \atop {x=\log_73}} \right.[/latex] Ответ  [latex]x=\log_72[/latex] или [latex]x=\log_73[/latex]