$$ \begin{array}{l}\log x+\log (x-6)=\log 27 \ \text { Solve the equation to find the solution set. Select the correct choice below and, if necessary, fill in the answer box to co } \ \text { A. The solution set is }\} \text {. } \ \text { (Simplify your answer. Use a comma to separate answers as needed.) } \ \text { B. There are infinitely many solutions. } \ \text { C. There is no solution. }\end{array} \text { } $$

Question

UpStudy AI · Accepted Answer

We start with the given equation:  
$$
\log x + \log (x-6) = \log 27
$$

**Step 1: Combine the logarithms**

Using the logarithm property  
$$
\log a + \log b = \log (ab),
$$  
we rewrite the equation as:  
$$
\log \left[x(x-6)\right] = \log 27.
$$

**Step 2: Equate the arguments**

Since the logarithm function is one-to-one, if  
$$
\log A = \log B,
$$  
then  
$$
A = B.
$$  
So we have:  
$$
x(x-6)=27.
$$

**Step 3: Form the quadratic equation**

Expanding the left side:  
$$
x^2 - 6x = 27.
$$  
Subtract 27 from both sides to set the equation to zero:  
$$
x^2 - 6x - 27 = 0.
$$

**Step 4: Solve the quadratic equation**

We calculate the discriminant:  
$$
\Delta = (-6)^2 - 4\cdot1\cdot(-27)=36+108=144.
$$  
Since $$\Delta=144$$, the solutions are given by:  
$$
x = \frac{-(-6) \pm \sqrt{144}}{2\cdot1} = \frac{6 \pm 12}{2}.
$$

Thus, we find two potential solutions:  
$$
x = \frac{6 + 12}{2} = 9 \quad \text{and} \quad x = \frac{6 - 12}{2} = -3.
$$

**Step 5: Check the domain restrictions**

For the original logarithmic expressions, the arguments must be positive:
- For $$\log x$$, we require $$x > 0$$.
- For $$\log (x-6)$$, we require $$x-6 > 0$$ or $$x > 6$$.

The value $$x = -3$$ does not satisfy $$x > 0$$ and $$x > 6$$.

Thus, the only valid solution is $$x = 9$$.

**Conclusion:**

The solution set is $$\{9\}$$.

The correct choice is A.
The solution set is {9}.

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