Solve: (a) 7^(4-16x) > 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, ∞) … [Read more...] about Algebra – Inequalities and Exponents

# GRE Inequalities

## Inequalities and Absolute Values

The solution set for the inequality |x+2| < |4x+1| (a) -3/5 < x < 1/3 (b) -1/3 < x < 3/5 (c) 0 < x < 1/3 (d) -1/3 < x < 1/3 (e) None of these Correct Answer: (E) Explanation Since the expressions on both sides of the inequality are under modulus, you can square on both sides without disturbing the inequality. (x+2)2 < … [Read more...] about Inequalities and Absolute Values