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

Explanation:

Step1: Isolate the natural - log term

Divide both sides of the equation $2\ln(x + 3)=0$ by 2. $\frac{2\ln(x + 3)}{2}=\frac{0}{2}$, so $\ln(x + 3)=0$.

Step2: Convert from logarithmic to exponential form

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

Step3: Solve for x

Subtract 3 from both sides of the equation $1=x + 3$. $x=1 - 3=-2$. We also need to check the domain of the original logarithmic function. The argument of the natural - log function $y = \ln u$ must be $u>0$. For $y=\ln(x + 3)$, we need $x+3>0$, or $x>-3$. $x=-2$ satisfies this condition. When we check $x=-4$, $\ln(-4 + 3)=\ln(-1)$ is undefined since the argument of the natural - log function must be positive.

Answer:

$x=-2$ (corresponding to the option where $x=-2$ is one of the values, which is the second option: $x=-2$ and $x = - 4$, but we discard $x=-4$ as it makes the logarithm undefined)