The base determines inequality direction. For base a > 1: $log_a$(f(x)) > $log_a$(g(x)) iff f(x) > g(x), with both f(x), g(x) > 0. For base 0 < a < 1: the inequality REVERSES — $log_a$(f(x)) > $log_a$(g(x)) iff f(x) < g(x), with both positive. When the base is variable, split into two cases. For $log_a$(f(x)) > k: if a > 1, then f(x) > $a^k$ (and f(x) > 0); if 0 < a < 1, then 0 < f(x) < $a^k$. The most common JEE pattern: solve $log_{1/2}$(expression) > 0, which means 0 < expression < 1 (since base < 1 and $log_{1/2}$(1) = 0). Always intersect the algebraic solution with the domain condition.