Решите уравнение _2(1-3x)=_4(5x-1).
Domain: 1-3x>0 and 5x-1>0 give x in (1/5, 1/3). Using log_4 t = (1/2) log_2 t, the equation is log_2(1-3x) = (1/2) log_2(5x-1), so (1-3x)^2 = 5x-1, i.e. 9x^2 - 11x + 2 = 0, with roots x=1 and x=2/9. Root x=1 fails the domain (1-3 = -2 < 0), so x=2/9 is the only solution. Check: 1-3x = 1/3, 5x-1 = 1/9, both logs equal -log_2(3).
x = 2/9