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

To solve the equation \(\log_{x-3}(x^2 - 9) = \log_{x-3}(x + 1) + 2\), we first need to consider the valid domain for the logarithms. The base of the logarithm, \(x - 3\), must be positive and cannot be 1. Thus, we have \(x - 3 > 0\) and \(x - 3 \neq 1\), which implies:

• \(x > 3\) (for the base to be positive) and 
• \(x \neq 4\) (to avoid the base being 1).

Additionally, the arguments of the logarithms must also be positive:

• \(x^2 - 9 > 0\), which yields \(x > 3\) or \(x < -3\).
• \(x + 1 > 0\), giving \(x > -1\).

Combining these conditions, we see that the valid range for \(x\) is \(x > 3\).

Returning to the original equation, we can use properties of logarithms to rewrite it as: \(\log_{x-3}(x^2 - 9) = \log_{x-3}((x + 1)(x - 3))\).

Thus, we simplify the equation using logarithmic properties: 

\[\log_{x-3}(x^2 - 9) = \log_{x-3}((x + 1)(x - 3)) + 2\]

.

Therefore, \(\log_{x-3}(x^2 - 9) - \log_{x-3}((x + 1)(x - 3)) = 2\).

Applying the quotient rule for logarithms gives: \(\log_{x-3}\left(\frac{x^2 - 9}{(x + 1)(x - 3)}\right) = 2\).

For this equality to hold, the quotient must be equal to the base raised to the power of 2:

• \(\frac{x^2 - 9}{(x + 1)(x - 3)} = (x - 3)^2\)
• Solving for \(x\) gives: \(\frac{(x - 3)(x + 3)}{(x + 1)(x - 3)} = (x - 3)^2\)
• Canceling the common factor \(x - 3\) (since \(x > 3\)): \(\frac{x + 3}{x + 1} = x - 3\)

Cross-multiplying gives us:

\[x + 3 = (x - 3)(x + 1)\]

Expanding the right-hand side:

\[x + 3 = x^2 - 3x + x - 3\]

Simplifying the equation:

\[x + 3 = x^2 - 2x - 3\]

Rearranging gives us the quadratic equation:

\[x^2 - 3x - 6 = 0\]

Solving this quadratic equation using the quadratic formula:

\[x = \frac{3 \pm \sqrt{9 + 4 \cdot 1 \cdot 6}}{2}\]

 

\[x = \frac{3 \pm \sqrt{33}}{2}\]

We have two potential solutions: \(\frac{3 + \sqrt{33}}{2}\) and \(\frac{3 - \sqrt{33}}{2}\). However, only \(\frac{3 + \sqrt{33}}{2}\) is greater than 3 and satisfies all the domain conditions.

Therefore, the sum of all possible real values of \(x\) is \(\frac{3 + \sqrt{33}}{2}\).

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: \[ \boxed{\dfrac{3 + \sqrt{33}}{2}}. \]