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

Question

Accepted Answer

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$({e^{2x}} - 4)(6{e^{2x}} - 5{e^x} + 1) = 0$

Let ${e^x} = t$

$\therefore$ $({t^2} - 4)(6{t^2} - 5t + 1) = 0$

$\Rightarrow ({t^2} - 4)(2t - 1)(3t - 1) = 0$

$\therefore$ t = 2, $-$2, ${1 \over 2}$, ${1 \over 3}$

$\therefore$ ${e^x} = 2 \Rightarrow x = \ln 2$

${e^x} = - 2$ (not possible)

${e^x} = {1 \over 2} \Rightarrow x = - \ln 2$

${e^x} = {1 \over 3} \Rightarrow x = - \ln 3$

$\therefore$ Sum of all real roots

$= \ln 2 - \ln 2 - \ln 3$

$= - \ln 3$

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The sum of all the real roots of the equation

$({e^{2x}} - 4)(6{e^{2x}} - 5{e^x} + 1) = 0$ is