1. **State the problem:** Solve the equation $$\ln(x + 70) + \ln x = \ln 71$$ for $x$. 2. **Recall the logarithm property:** The sum of logarithms is the logarithm of the product: $$\ln a + \ln b = \ln(ab)$$ 3. **Apply the property:** $$\ln(x + 70) + \ln x = \ln((x + 70) \cdot x) = \ln(71)$$ 4. **Set the arguments equal:** Since $\ln A = \ln B$ implies $A = B$, we have: $$(x + 70) \cdot x = 71$$ 5. **Write the quadratic equation:** $$x^2 + 70x = 71$$ 6. **Bring all terms to one side:** $$x^2 + 70x - 71 = 0$$ 7. **Use the quadratic formula:** For $ax^2 + bx + c = 0$, $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ Here, $a=1$, $b=70$, $c=-71$. 8. **Calculate the discriminant:** $$\Delta = b^2 - 4ac = 70^2 - 4 \cdot 1 \cdot (-71) = 4900 + 284 = 5184$$ 9. **Calculate the square root:** $$\sqrt{5184} = 72$$ 10. **Find the roots:** $$x = \frac{-70 \pm 72}{2}$$ - For $+$: $$x = \frac{-70 + 72}{2} = \frac{2}{2} = 1$$ - For $-$: $$x = \frac{-70 - 72}{2} = \frac{-142}{2} = -71$$ 11. **Check domain restrictions:** Since $\ln(x)$ and $\ln(x+70)$ require $x > 0$ and $x+70 > 0$, $x = -71$ is invalid. 12. **Final answer:** $$\boxed{x = 1}$$