Log (x) 17+log (17) x<=2

Решить неравенство

Log17 (x) + logx (17) < = 2

x > 0, x! = 1

logx (17) = 1/log17 (x)

log17 (x) + 1/log17 (x) < = 2

log17 (x) = a

a + 1/a < = 2

a + 1/a - 2 < = 0

a^2 + 1 - 2a < = 0

a1 + a2 = 2

a1*a2 = 1

a = 1

- 1 +

a < = 1

log17 (x) < = 1

x < = 17

c учетом x > 0, x! = 1 получаем ответ:

0 < x < = 17, x! = 1