Solve:
(a) 7^(4-16x) > 1
(b) 2/(x-1) < 1
(b) 2/(x-1) < 1
(a) For any a^b > 1, if a > 1 then b should be greater than 0.
Hence 4 – 16x > 0
Therefore, x<1/4.
(b) 2/(x-1) < 1
2/(x-1) – 1< 0
[2-(x-1)]/(x-1) < 0
(3-x)/(x-1) < 0
(x-3)/(x-1) > 0
Hence ‘x’ lies outside the roots ‘1’ and ‘3’.
Therefore, x € (-∞ , 1) U (3, ∞)
Questions, answers, comments welcome