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