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

Using the Solving Logarithmic Equations knowledge point

$$
LATEXBLOCK0
$$

Convert to exponential form

Using the Solving Logarithmic Equations knowledge point

$$
LATEXBLOCK1
$$

Solve for x

Using the Solving Logarithmic Equations knowledge point

$$
LATEXBLOCK2
$$

Analyze the given options

Using the Solving Logarithmic Equations knowledge point
The algebraic solution is \(x = -2\). However, looking closely at the provided multiple-choice options, none of them list \(x = -2\) as the sole solution. Instead, they list pairs of potential solutions:

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

The only option containing the correct mathematical solution \(x = -2\) is the second option: "\(x = -2\) and \(x = -4\)". In many algebraic contexts, solving an equation rewritten using logarithmic properties (such as \(\ln(x+3)^2 = 0\)) can introduce extraneous solutions like \(x = -4\). Let's verify this path:

$$
LATEXBLOCK3
$$

Thus, the potential solutions derived from applying the power rule of logarithms are \(x = -2\) and \(x = -4\).

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 \(x = 4\)