what are the potential solutions to the equation below? 2ln(x + 3)=0\nx=-3 and x=-4\nx=-2 and x=-4\nx=2 and…

Explanation:

Step1: Isolate the natural - logarithm term

Divide both sides of the equation $2\ln(x + 3)=0$ by 2. We get $\ln(x + 3)=0$.

Step2: Convert the logarithmic equation to an exponential equation

Recall that if $\ln a=b$, then $e^{b}=a$. Since $\ln(x + 3)=0$ and $e^{0}=1$, we have $x + 3=1$.

Step3: Solve for x

Subtract 3 from both sides of the equation $x + 3=1$. So $x=1 - 3=-2$. We also need to check the domain of the original logarithmic function. The argument of the natural - logarithm $\ln(x + 3)$ must be positive, i.e., $x+3>0$ or $x>-3$. $x = - 2$ satisfies this condition. When we check $x=-4$, $\ln(-4 + 3)=\ln(-1)$ is undefined in the real - number system.

Answer:

The solution is $x=-2$, so the answer is the option with $x=-2$ and $x=-4$ (but we discard $x = - 4$ as it makes the logarithm undefined), which means the correct option is the one with $x=-2$ among the given choices. In the multiple - choice context, the correct option is: $x=-2$ and $x=-4$ (where $x=-2$ is the valid solution and $x=-4$ is an extraneous solution due to the domain of the logarithm).