MathematicsEasy129×since 2002Q5100The sum of all the real roots of the equation (e2x−4)(6e2x−5ex+1)=0({e^{2x}} - 4)(6{e^{2x}} - 5{e^x} + 1) = 0(e2x−4)(6e2x−5ex+1)=0 isAloge3{\log _e}3loge3B−loge3- {\log _e}3−loge3Cloge6{\log _e}6loge6D−loge6- {\log _e}6−loge6Check answerSkip