Let $x=e^t\Rightarrow t={\log }_{e}x$

Then, the given equation reduces to $x^4-10x^3+29x^2-22x+4=0$

Now, product of roots taken all at a time of a biquadratic equation is $\frac{\mathrm{constant} \mathrm{term}}{\mathrm{coefficient} \mathrm{of} x^2}$.

Let $x_1, x_2, x_3$ and $x_4$ are the four roots of $x^4-10x^3+29x^2-22x+4=0$, then $x_1{x}_2{x}_3{x}_4=\frac{4}{1}=4$.

Taking $\log$ both sides, we get ${\log }_e\left(x_1{x}_2{x}_3{x}_4\right)={\log }_{e}4$

$\Rightarrow {\log }_e{x}_1+{\log }_e{x}_2+{\log }_e{x}_3+{\log }_e{x}_4=2{\log }_{e}2$

$\Rightarrow t_1+t_2+t_3+t_4=2{\log }_{e}2$

Where $t_1, t_2, t_3$ and $t_4$ are the roots of the equation $e^{4t}-10e^{3t}+29e^{2t}-22e^t+4=0$.

$$Hence, sum of roots is $2{\log }_{e}2$.