Question

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

Explanation:

Step1: Divide both sides by 2

To isolate the natural logarithm term, we divide both sides of the equation \(2\ln(x + 3)=0\) by 2. This gives us \(\ln(x + 3)=\frac{0}{2} = 0\).

Step2: Convert logarithmic to exponential form

Recall that if \(\ln(a)=b\), then \(a = e^{b}\). Since \(\ln(x + 3)=0\), we can rewrite this in exponential form as \(x + 3=e^{0}\).

Step3: Simplify \(e^{0}\) and solve for \(x\)

We know that any non - zero number to the power of 0 is 1, so \(e^{0}=1\). Then our equation becomes \(x + 3=1\). Subtracting 3 from both sides, we get \(x=1 - 3=-2\). We also need to check the domain of the original logarithmic function. The argument of \(\ln(x + 3)\) must be greater than 0, i.e., \(x+3>0\) or \(x>- 3\). Now we check the other values from the options. For \(x=-4\), \(x + 3=-1<0\) (not in domain), for \(x = - 3\), \(x + 3 = 0\) (not in domain), for \(x = 2\), let's check the original equation: \(2\ln(2 + 3)=2\ln(5)
eq0\), for \(x = 4\), \(2\ln(4 + 3)=2\ln(7)
eq0\). But wait, maybe we made a mistake in considering the options. Wait, let's re - evaluate the options. Wait, the options have two values, maybe there was a miscalculation. Wait, no, the equation \(2\ln(x + 3)=0\) has only one solution \(x=-2\), but the options are pairs. Wait, maybe there is a mistake in the problem or in my approach? Wait, no, let's check the options again. Wait, maybe the original equation was supposed to be \(2\ln|x + 3|=0\)? If that's the case, then \(\ln|x + 3| = 0\), \(|x + 3|=e^{0}=1\), so \(x+3 = 1\) or \(x + 3=-1\). Solving \(x + 3=1\) gives \(x=-2\), solving \(x + 3=-1\) gives \(x=-4\). Now we check the domain for the absolute - value logarithmic function. The argument of \(\ln|x + 3|\) is \(|x + 3|\), which is always positive for \(x
eq - 3\). So if the equation was \(2\ln|x + 3|=0\) (maybe a typo in the original problem, missing the absolute value), then the solutions are \(x=-2\) and \(x=-4\), which is the second option.

Answer:

B. \(x = - 2\) and \(x=-4\)