To solve the equation \( \log x + \log (x + 3) = 1 \), we can use the properties of logarithms. Specifically, we use the property that states \( \log a + \log b = \log (ab) \).

1. Combine the logarithms:
   \[
   \log (x(x + 3)) = 1
   \]

2. Exponentiate both sides to eliminate the logarithm:
   \[
   x(x + 3) = 10^1
   \]
   \[
   x(x + 3) = 10
   \]

3. Expand the left side:
   \[
   x^2 + 3x - 10 = 0
   \]

4. Now, we can solve this quadratic equation using the quadratic formula:
   \[
   x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
   \]
   where \( a = 1, b = 3, c = -10 \).

5. Calculate the discriminant:
   \[
   b^2 - 4ac = 3^2 - 4(1)(-10) = 9 + 40 = 49
   \]

6. Substitute into the quadratic formula:
   \[
   x = \frac{-3 \pm \sqrt{49}}{2(1)} = \frac{-3 \pm 7}{2}
   \]

7. This gives us two solutions:
   \[
   x = \frac{4}{2} = 2 \quad \text{and} \quad x = \frac{-10}{2} = -5
   \]

Since \( x \) must be positive in the context of logarithms, we discard \( x = -5 \).

Thus, the solution is:
\[
\boxed{2}
\]

### Would you like details on any step? 

Here are some related questions you might consider:
1. How do logarithmic properties help in solving equations?
2. What are the conditions under which a logarithmic function is defined?
3. Can you provide examples of other equations involving logarithms?
4. What happens if we consider \( \log x + \log (x + 3) = k \) for other values of \( k \)?
5. How can quadratic equations be solved using factoring? 

**Tip:** Always check your solutions in the original equation, especially when dealing with logarithms, to avoid extraneous solutions.

Explore the solution to the logarithmic equation Log(x) + Log(x + 3) = 1 with a step-by-step approach. This guide covers key logarithmic properties and the application of the quadratic formula, making it suitable for students in Grades 9-12.

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