Question

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 (x = -3)
(x = 2) and (x = 4)

Explanation:

Isolate the logarithmic term

$$
LATEXBLOCK0
$$

Convert to exponential form

$$
LATEXBLOCK1
$$

Solve for x and verify

$$
LATEXBLOCK2
$$

Since \(x = -2\) makes the argument \(x + 3 = 1 > 0\), it is a valid solution. Looking at the options, the second option contains \(x = -2\).

Answer:

• (A) \(x = -3\) and \(x = -4\)
• (B) \(x = -2\) and \(x = -4\) (Correct answer)
• (C) \(x = 2\) and \(x = -3\)
• (D) \(x = 2\) and \(x = 4\)