49^x - 6*7^x + 5 > 0
Пусть 7^x = t, тогда
t^2 - 6t + 5 > 0
t^2 - 6t + 5 = 0
D = 36 - 20 = 16
t1 = ( 6 + 4)/2 = 10/2 = 5
t2 = (6 - 4)/2 = 2/2 = 1
Обратная замена
7^x = 5
x = log 7 (5)
7^x = 1
x = 0
+ - +
----------------- 0 ----------------- log 7 (5) ------------> x
x ∈ ( - ∞; 0 ) ∪ (log 7 (5) ; + ∞)