Q. which expression is equivalent to \log_{3}(x + 4)?

Problem

Find expressions equivalent to \( \log_3(x+4) \). Provide detailed, step-by-step explanations and domain conditions.

1. Determine the domain of the original expression.

   The logarithm \( \log_3(x+4) \) is defined when its argument is positive. Thus

   \( x+4>0 \).

   So the domain is

   \( x>-4 \).

2. Change-of-base formula (using natural logarithm).

   The change-of-base formula for logarithms states that for positive a not equal to 1 and positive b,

   \( \log_a b = \dfrac{\ln b}{\ln a} \).

   Applying this with \( a=3 \) and \( b=x+4 \) gives

   \( \log_3(x+4) = \dfrac{\ln(x+4)}{\ln 3} \).

   Domain condition: the same as the original, \( x>-4 \), because \( \ln(x+4) \) requires \( x+4>0 \).

3. Factorization as a product: use \( x+4 = x\left(1+\dfrac{4}{x}\right) \).

   Start with the factorization

   \( x+4 = x\Bigl(1+\dfrac{4}{x}\Bigr) \).

   Apply the logarithm product rule \( \log_3(AB)=\log_3 A + \log_3 B \) to obtain

   \( \log_3(x+4) = \log_3 x + \log_3\Bigl(1+\dfrac{4}{x}\Bigr). \)

   Domain condition for this form: both \( \log_3 x \) and \( \log_3\bigl(1+\dfrac{4}{x}\bigr) \) must be defined. That requires \( x>0 \). If \( x>0 \) then \( 1+\dfrac{4}{x}>1>0 \), so this decomposition is valid for \( x>0 \). Note that this is a stricter domain than the original \( x>-4 \).

4. Factorization using 4: use \( x+4 = 4\left(1+\dfrac{x}{4}\right) \).

   Write

   \( x+4 = 4\Bigl(1+\dfrac{x}{4}\Bigr) \).

   Apply the logarithm product rule to get

   \( \log_3(x+4) = \log_3 4 + \log_3\Bigl(1+\dfrac{x}{4}\Bigr). \)

   Domain condition: \( 1+\dfrac{x}{4}>0 \) which is equivalent to \( x>-4 \). This matches the original domain.

5. Change-of-base using common logarithm (base 10).

   Using base-10 logs, the change-of-base formula gives

   \( \log_3(x+4) = \dfrac{\log(x+4)}{\log 3} \).

   Domain: \( x>-4 \).

Summary — equivalent expressions (with domains):

• \( \log_3(x+4) = \dfrac{\ln(x+4)}{\ln 3} \), valid for \( x>-4 \).
• \( \log_3(x+4) = \dfrac{\log(x+4)}{\log 3} \), valid for \( x>-4 \).
• \( \log_3(x+4) = \log_3 4 + \log_3\Bigl(1+\dfrac{x}{4}\Bigr) \), valid for \( x>-4 \).
• \( \log_3(x+4) = \log_3 x + \log_3\Bigl(1+\dfrac{4}{x}\Bigr) \), valid for \( x>0 \) (this has the stricter domain requirement).