Question

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

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

Answer 1

Answer: (B)

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