log pomocyyy: log_x 4<2

Janek191: x > 0 i x ≠ 1 log_x 4 < 2 log_x 4 < 2 log_x x log_x 4 < log_x x² 1) x ∊( 0; 1) wtedy 4 > x² x² < 4 x ∊ ( − 2; 2) więc x ∊ ( 0; 1) −−−−−−−−−−−− 2) x > 1 wtedy 4 < x² x² > 4 x ∊ ( − ∞ ; − 2)∪ ( 2; +∞) więc x ∊ (2; +∞) −−−−−−−− Odp. x ∊ ( 0; 1) ∪ ( 2 : +∞) ======================