Question

Решить 49^x-5*7^x+6=0.

Answer 1

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

Ответ 
$x=log_72$ или $x=log_73$