Question

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

• A. $log\;e^3$
• B. $–log\;e^3$
• C. $log\;e^6$
• D. $–log\;e^6$

Answer 1

Answer: (B)

$–log\;e^3$

Answer 2

Given equation : $(e^{2x} – 4)(6e^{2x} – 5e^x + 1) = 0$

$⇒ e^{2x} – 4 = 0 \;or \;6e^{2x} – 5e^x + 1 = 0$

$⇒ e^{2x} = 4 \;or\; 6(e^x)^2 – 3e^x – 2e^x + 1 = 0$

$⇒ 2x = ln4 \;or (3e^x – 1)(2e^x – 1) = 0$

$⇒ x = In2 \;or\; e^x = \frac{1}{3}\; or\; e^x = \frac{1}{2}$

else, $x = ln\frac{1}{3}, -ln2$

Sum of all real roots = $ln2 – ln3 – ln2$

= $–ln3$

Hence, the correct option is (B): $–log\;e^3$