Question

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

• A. ${\log }_{e}3$
• B. $-{\log }_{e}3$
• C. ${\log }_{e}6$
• D. $-{\log }_{e}6$

Answer 1

Answer: (B)

$\left(e^{2x}-4\right)\left(6e^{2x}-3e^x-2e^x+1\right)=0$
$\left(e^{2x}-4\right)\left(3e^x-1\right)\left(2e^x-1\right)=0$
$e^{2x}=4$ or $e^x=\frac{1}{3}$ or $e^x=\frac{1}{2}$
$\Rightarrow$ sum of real roots $=\frac{1}{2}\ln 4+\ln \frac{1}{3}+\ln \frac{1}{2}$
$=-\ln 3$