JEE Main 202224 Jun, Evening shiftMathematicsEasyComplex Numbers and Quadratic Equations264× since 2002The 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 isAOption A. loge3{\log _e}3loge3BOption B. −loge3- {\log _e}3−loge3COption C. loge6{\log _e}6loge6DOption D. −loge6- {\log _e}6−loge6CheckSkip · show solution
