Question

The sum of all possible real values of $x$ for which \[ \log_{x-3}(x^2 - 9) = \log_{x-3}(x + 1) + 2, \] is

• A. \(-3\)
• B. \(\sqrt{33}\)
• C. \(\dfrac{3 + \sqrt{33}}{2}\)
• D. \(3\)

Hint:

In logarithmic equations:
Always check the domain first: base $> 0$, base $\neq 1$, and argument $> 0$.
When the bases are the same, combine logs using properties like $\log_b A - \log_b B = \log_b \left(\dfrac{A}{B}\right)$, then convert to exponential form.
Don’t forget to discard any solutions that fall outside the domain constraints.

Answer 1

The Correct Option is C

Approach: First nail the domain (logs are picky about base and arguments), then turn the "\(+2\)" into \(\log_{x-3}(x-3)^2\) so both sides become a single log and the bases cancel.

Step 1 (domain): Base \(x-3>0\) and \(x-3\neq1\) give \(x>3,\ x\neq4.\) Arguments: \(x^2-9>0\) and \(x+1>0\) are both satisfied once \(x>3.\) So the legal zone is \(x>3.\)

Step 2: Write \(2=\log_{x-3}(x-3)^2.\) The equation becomes
\[\log_{x-3}(x^2-9)=\log_{x-3}\big((x+1)(x-3)^2\big).\]

Step 3: Equal logs (same base) means equal arguments:
\[x^2-9=(x+1)(x-3)^2.\]
Since \(x^2-9=(x-3)(x+3)\) and \(x>3\) lets us cancel one \((x-3)\):
\[x+3=(x+1)(x-3).\]

Step 4: Expand: \(x+3=x^2-2x-3\Rightarrow x^2-3x-6=0.\) By the quadratic formula \(x=\dfrac{3\pm\sqrt{9+24}}{2}=\dfrac{3\pm\sqrt{33}}{2}.\)

Step 5 (domain filter): \(\dfrac{3-\sqrt{33}}{2}\approx-1.37\) fails \(x>3\); only \(\dfrac{3+\sqrt{33}}{2}\approx4.37\) survives. (The cancelled \(x=3\) is also rejected \(-\) it makes the base \(0\).) With a single valid root, the "sum of all possible real values" is that root itself.

Final answer: \(\dfrac{3+\sqrt{33}}{2}.\)

Answer 2

Step 1: Determine the domain. The base of the logarithm is \(x - 3\), so we need: \[ x - 3>0 \quad \Rightarrow \quad x>3, \] and \[ x - 3 \neq 1 \quad \Rightarrow \quad x \neq 4. \] Also, the arguments of the logarithms must be positive: \[ x^2 - 9>0 \quad \Rightarrow \quad x>3 \text{ or } x<-3, \] \[ x + 1>0 \quad \Rightarrow \quad x>-1. \] Combining all conditions: \[ x>3, \quad x \neq 4. \] 
Step 2: Simplify the equation using log properties. Given: \[ \log_{x-3}(x^2 - 9) = \log_{x-3}(x + 1) + 2. \] Bring logs together: \[ \log_{x-3}(x^2 - 9) - \log_{x-3}(x + 1) = 2. \] Using the property \(\log_b A - \log_b B = \log_b \left(\dfrac{A}{B}\right)\): \[ \log_{x-3}\left(\frac{x^2 - 9}{x + 1}\right) = 2. \] Factor: \[ x^2 - 9 = (x - 3)(x + 3), \quad \Rightarrow \quad \frac{x^2 - 9}{x + 1} = \frac{(x - 3)(x + 3)}{x + 1}. \] So: \[ \log_{x-3}\left(\frac{(x - 3)(x + 3)}{x + 1}\right) = 2. \] 
Step 3: Convert the logarithmic equation to exponential form. \[ \log_{x-3}\left(\frac{(x - 3)(x + 3)}{x + 1}\right) = 2 \quad \Rightarrow \quad \frac{(x - 3)(x + 3)}{x + 1} = (x - 3)^2. \] Since \(x>3\), we have \(x - 3 \neq 0\), so we can safely multiply and divide. \[ \frac{(x - 3)(x + 3)}{x + 1} = (x - 3)^2 \quad \Rightarrow \quad (x - 3)(x + 3) = (x - 3)^2 (x + 1). \] Divide both sides by \(x - 3\): \[ x + 3 = (x - 3)(x + 1). \] Expand the right-hand side: \[ x + 3 = x^2 - 2x - 3. \] Rearrange: \[ 0 = x^2 - 2x - 3 - x - 3 = x^2 - 3x - 6. \] 
Step 4: Solve the quadratic and apply the domain. \[ x^2 - 3x - 6 = 0 \quad \Rightarrow \quad x = \frac{3 \pm \sqrt{9 + 24}}{2} = \frac{3 \pm \sqrt{33}}{2}. \] We must satisfy \(x>3\). \[ \frac{3 - \sqrt{33}}{2}<0 \quad (\text{reject}), \quad \frac{3 + \sqrt{33}}{2}>3 \quad (\text{accept}). \] Hence the only valid solution is: \[ x = \frac{3 + \sqrt{33}}{2}. \] Since the question asks for the sum of all possible real values of \(x\), the sum is just this value: \[ \boxed{\dfrac{3 + \sqrt{33}}{2}}. \]