1. **State the problem:** Solve for $x$ in the equation $$(6 - \sqrt{35})^x + (6 + \sqrt{35})^x = 142.$$\n\n2. **Recognize the form:** Let $a = 6 - \sqrt{35}$ and $b = 6 + \sqrt{35}$. The equation becomes $$a^x + b^x = 142.$$\n\n3. **Important property:** Note that $a$ and $b$ are conjugates, and their product is $$ab = (6 - \sqrt{35})(6 + \sqrt{35}) = 6^2 - (\sqrt{35})^2 = 36 - 35 = 1.$$\n\n4. **Check for integer $x$:** Since $ab=1$, $b = \frac{1}{a}$. The equation can be rewritten as $$a^x + \left(\frac{1}{a}\right)^x = 142,$$ or $$a^x + a^{-x} = 142.$$\n\n5. **Set $y = a^x$:** Then the equation is $$y + \frac{1}{y} = 142.$$ Multiply both sides by $y$ to get $$y^2 + 1 = 142y.$$\n\n6. **Rewrite as quadratic:** $$y^2 - 142y + 1 = 0.$$\n\n7. **Solve quadratic for $y$:** Using the quadratic formula $$y = \frac{142 \pm \sqrt{142^2 - 4}}{2} = \frac{142 \pm \sqrt{20164 - 4}}{2} = \frac{142 \pm \sqrt{20160}}{2}.$$\n\n8. **Simplify the square root:** $$\sqrt{20160} = \sqrt{64 \times 315} = 8 \sqrt{315}.$$ So $$y = \frac{142 \pm 8 \sqrt{315}}{2} = 71 \pm 4 \sqrt{315}.$$\n\n9. **Recall $y = a^x$ and $a = 6 - \sqrt{35} < 1$:** Since $a < 1$, $a^x$ decreases as $x$ increases. The two possible values for $y$ are $$71 + 4 \sqrt{315} > 1$$ and $$71 - 4 \sqrt{315}.$$\n\n10. **Check which $y$ is valid:** Since $a^x$ is positive and less than 1 for positive $x$, the value $71 + 4 \sqrt{315}$ is too large. The smaller value $$71 - 4 \sqrt{315}$$ is positive and less than 1, so $$a^x = 71 - 4 \sqrt{315}.$$\n\n11. **Solve for $x$:** $$x = \frac{\ln\left(71 - 4 \sqrt{315}\right)}{\ln\left(6 - \sqrt{35}\right)}.$$\n\n**Final answer:** $$\boxed{x = \frac{\ln\left(71 - 4 \sqrt{315}\right)}{\ln\left(6 - \sqrt{35}\right)}}.$$