The sum of all the real roots of the equation (e^2x-4)(6e^2x-5e^x+1)=0 is

Question

The sum of all the real roots of the equation $(e^{2x}-4)(6e^{2x}-5e^x+1)=0$ is

Accepted Answer

Factor 1: $e^{2x}=4 \implies x=\ln2$ (a valid real solution).

Factor 2: Let $t=e^x$. $6t^2-5t+1=0 \implies t=\dfrac12$ or $\dfrac13$, giving $x=-\ln2$ or $x=-\ln3$.

$$\text{Sum} = \ln2+(-\ln2)+(-\ln3) = -\ln3.$$

Solutions · 1