1. **Problem 1:** Solve the equation $$\ln(x + 70) + \ln x = \ln 71$$. 2. **Problem 2:** Solve the equation $$2 \cdot 6^{m+2} + 4 = 44$$. --- ### Problem 1: Solve $$\ln(x + 70) + \ln x = \ln 71$$ 1. Recall the logarithm property: $$\ln a + \ln b = \ln(ab)$$. 2. Apply this property to combine the left side: $$\ln((x + 70) \cdot x) = \ln 71$$. 3. Since $$\ln A = \ln B$$ implies $$A = B$$, we have: $$x(x + 70) = 71$$. 4. Expand the left side: $$x^2 + 70x = 71$$. 5. Rearrange to standard quadratic form: $$x^2 + 70x - 71 = 0$$. 6. Use the quadratic formula: $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ where $$a=1$$, $$b=70$$, $$c=-71$$. 7. Calculate the discriminant: $$\Delta = 70^2 - 4 \cdot 1 \cdot (-71) = 4900 + 284 = 5184$$. 8. Calculate the square root: $$\sqrt{5184} = 72$$. 9. Find the roots: $$x = \frac{-70 \pm 72}{2}$$. 10. Calculate each root: - $$x_1 = \frac{-70 + 72}{2} = \frac{2}{2} = 1$$ - $$x_2 = \frac{-70 - 72}{2} = \frac{-142}{2} = -71$$ 11. Check domain restrictions for logarithms: $$x > 0$$ and $$x + 70 > 0$$. 12. Discard $$x = -71$$ because it is not in the domain. **Final solution for Problem 1:** $$x = 1$$. --- ### Problem 2: Solve $$2 \cdot 6^{m+2} + 4 = 44$$ 1. Subtract 4 from both sides: $$2 \cdot 6^{m+2} = 40$$. 2. Divide both sides by 2: $$6^{m+2} = 20$$. 3. Rewrite the exponent: $$6^{m} \cdot 6^{2} = 20$$. 4. Since $$6^2 = 36$$, we have: $$36 \cdot 6^{m} = 20$$. 5. Divide both sides by 36: $$6^{m} = \frac{20}{36} = \frac{5}{9}$$. 6. Take the natural logarithm of both sides: $$\ln(6^{m}) = \ln\left(\frac{5}{9}\right)$$. 7. Use the logarithm power rule: $$m \ln 6 = \ln\left(\frac{5}{9}\right)$$. 8. Solve for $$m$$: $$m = \frac{\ln\left(\frac{5}{9}\right)}{\ln 6}$$. **Final solution for Problem 2:** $$m = \frac{\ln\left(\frac{5}{9}\right)}{\ln 6}$$.