Use Newton's method to find an approximate solution of $\ln (x)=10-x$. Start with $x_{0}=6$ and find $x_{2}$. \[ \mathrm{x}_{2}= \] (Do not round until the final answer. Then round to six decimal places as needed.)

Step 1 ：\(x_0 = 6\)

Step 2 ：\(x_1 = x_0 - \frac{f(x_0)}{f'(x_0)} = 6 - \frac{\ln(6) + 6 - 10}{1/6 + 1} = 6 - \frac{\ln(6) - 4}{7/6} = 6 - 6*(\ln(6) - 4)*6/7 = 6 - 6*(0.1823215567939546 - 4)/7 = 6 - 6*(-3.8176784432060454)/7 = 6 + 3.263440239317189 = 9.263440239317189\)

Step 3 ：\(x_2 = x_1 - \frac{f(x_1)}{f'(x_1)} = 9.263440239317189 - \frac{\ln(9.263440239317189) + 9.263440239317189 - 10}{1/9.263440239317189 + 1} = 9.263440239317189 - \frac{2.226508669456576 + 9.263440239317189 - 10}{0.107961425681448 + 1} = 9.263440239317189 - \frac{1.489948908773765}{1.107961425681448} = 9.263440239317189 - 1.344498680711018 = 7.918941558606171\)

Step 4 ：\(\boxed{x_2 = 7.918942}\)