To solve the equation

$\log x + \log (x + 3) = 1,$

we can proceed with the following steps:

1. Combine the logarithmic terms:
   Using the property $\log a + \log b = \log (a \cdot b)$, we rewrite the left side as a single logarithm:

   $\log (x \cdot (x + 3)) = 1.$

   Simplifying inside the logarithm:

   $\log (x^2 + 3x) = 1.$

2. Rewrite in exponential form:
   Since $\log (x^2 + 3x) = 1$, we can rewrite this equation in exponential form as:

   $x^2 + 3x = 10^1.$

   Simplifying the right side:

   $x^2 + 3x = 10.$

3. Rewrite as a standard quadratic equation:
   Move all terms to one side to form a quadratic equation:

   $x^2 + 3x - 10 = 0.$

4. Solve the quadratic equation:
   We can use the quadratic formula, $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, where $a = 1$, $b = 3$, and $c = -10$:

   $x = \frac{-3 \pm \sqrt{3^2 - 4 \cdot 1 \cdot (-10)}}{2 \cdot 1}.$

   Simplify under the square root:

   $x = \frac{-3 \pm \sqrt{9 + 40}}{2}.$

   $x = \frac{-3 \pm \sqrt{49}}{2}.$

   $x = \frac{-3 \pm 7}{2}.$

   This gives two solutions:

   $x = \frac{-3 + 7}{2} = 2 \quad \text{and} \quad x = \frac{-3 - 7}{2} = -5.$

5. Check for valid solutions:
   Since $x = -5$ would make $\log x$ undefined (we cannot take the logarithm of a negative number), we discard $x = -5$. Thus, the only valid solution is:

   $x = 2.$

So, the solution set is: $\{2\}.$

Would you like further details or have any questions?

Here are 5 related questions for further practice:

1. Solve $\log (x + 2) + \log (x + 4) = 2$.
2. If $\log x + \log (x - 5) = 2$, find $x$.
3. Solve the equation $\log (2x) + \log (x + 1) = 1$.
4. Determine the solution to $\log (x) + \log (x + 6) = 2$.
5. Solve $\log (x + 3) + \log (x + 5) = 1$.

Tip: Always check if the values satisfy the domain of the logarithmic function, especially in cases where the solution includes negative or zero values.