The sum of the roots of the equation e^{4t} - | vedclass

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(B) Let $x = e^t$,which implies $t = \log_e x$. Then,the given equation reduces to $x^4 - 10x^3 + 29x^2 - 22x + 4 = 0$. Let $x_1, x_2, x_3, x_4$ be th

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$2 \log_e 2$

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(B) Let $x = e^t$,which implies $t = \log_e x$. Then,the given equation reduces to $x^4 - 10x^3 + 29x^2 - 22x + 4 = 0$. Let $x_1, x_2, x_3, x_4$ be the roots of this biquadratic equation. By Vieta's formulas,the product of the roots is $x_1 x_2 x_3 x_4 = \frac{	ext{constant term}}{	ext{coefficient of } x^4} = \frac{4}{1} = 4$. Taking the natural logarithm on both sides,we get $\log_e(x_1 x_2 x_3 x_4) = \log_e 4$. Using the property $\log(ab) = \log a + \log b$,we have $\log_e x_1 + \log_e x_2 + \log_e x_3 + \log_e x_4 = \log_e(2^2) = 2 \log_e 2$. Since $t_i = \log_e x_i$ are the roots of the original equation in $t$,the sum of the roots is $t_1 + t_2 + t_3 + t_4 = 2 \log_e 2$.

The sum of the roots of the equation $e^{4t} - 10e^{3t} + 29e^{2t} - 22e^t + 4 = 0$ is

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