T trandoan New member Xu 0 16/12/10 #1 {4}^{x}-{2}^{x}={16}^{y}-{4}^{y} {log}_{3}x-{log}_{2}\left(2y-1 \right)=0
coconvuive12 New member Xu 0 16/12/10 #2 \[\left {{4}^{x}-{2}^{x}={16}^{y}-{4}^{y}\\ {{log}_{3}x-{log}_2}\left(2y-1 \right)=0\]