Step 1: Apply the logarithm property $\ln a + \ln b = \ln(ab)$. The given equation is: $$\ln(x+3) + \ln(x+5) = \ln15$$ Using the property, we combine the terms on the left side: $$\ln((x+3)(x+5)) = \ln15$$ Step 2: Equate the arguments of the logarithms. Since the natural logarithm is a one-to-one function, if $\ln A = \ln B$, then $A=B$. $$(x+3)(x+5) = 15$$ Step 3: Expand and solve the quadratic equation. Expand the left side: $$x^2 + 5x + 3x + 15 = 15$$ $$x^2 + 8x + 15 = 15$$ Subtract 15 from both sides: $$x^2 + 8x = 0$$ Factor out $x$: $$x(x+8) = 0$$ This gives two potential solutions: $$x = 0 \quad \text{or} \quad x+8 = 0 \implies x = -8$$ Step 4: Check the validity of the solutions based on the domain of the natural logarithm. For $\ln(x+3)$ to be defined, $x+3 > 0 \implies x > -3$. For $\ln(x+5)$ to be defined, $x+5 > 0 \implies x > -5$. Both conditions must be satisfied, so we need $x > -3$. Let's check our potential solutions: • For $x=0$: $0 > -3$. This condition is satisfied. Substituting $x=0$ into the original equation: $\ln(0+3) + \ln(0+5) = \ln3 + \ln5 = \ln(3 \times 5) = \ln15$. This is correct. So, $x=0$ is a valid solution. • For $x=-8$: $-8 > -3$. This condition is NOT satisfied. Substituting $x=-8$ into the original equation would give $\ln(-8+3) = \ln(-5)$ and $\ln(-8+5) = \ln(-3)$, which are undefined in the set of real numbers. So, $x=-8$ is not a valid solution. Therefore, the only valid solution is $x=0$. Step 5: Select the correct option. Based on our analysis, the correct option is A. The final answer is $\boxed{\text{A) x=0 only}}$. Send me the next one 📸