The sum of all the real roots of the equation | vedclass

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(B) Given equation: $(e^{2x} - 4)(6e^{2x} - 5e^x + 1) = 0$Factorizing the quadratic part: $6e^{2x} - 5e^x + 1 = 6e^{2x} - 3e^x - 2e^x + 1 = 3e^x(2e^x

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$-\log_{e} 3$

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(B) Given equation: $(e^{2x} - 4)(6e^{2x} - 5e^x + 1) = 0$Factorizing the quadratic part: $6e^{2x} - 5e^x + 1 = 6e^{2x} - 3e^x - 2e^x + 1 = 3e^x(2e^x - 1) - 1(2e^x - 1) = (3e^x - 1)(2e^x - 1) = 0$So,the equation becomes: $(e^{2x} - 4)(3e^x - 1)(2e^x - 1) = 0$This gives three possible cases:$1) e^{2x} = 4$ $\Rightarrow 2x = \ln 4$ $\Rightarrow x = \frac{1}{2} \ln 4 = \ln 2$$2) 3e^x = 1$ $\Rightarrow e^x = \frac{1}{3}$ $\Rightarrow x = \ln(\frac{1}{3}) = -\ln 3$$3) 2e^x = 1$ $\Rightarrow e^x = \frac{1}{2}$ $\Rightarrow x = \ln(\frac{1}{2}) = -\ln 2$Sum of all real roots = $\ln 2 + (-\ln 3) + (-\ln 2) = -\ln 3$.

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

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