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The equation you provided is:

\[
\log_5{x} + \log_5{(x+1)} = 1 + \log_5{(x+12)}
\]

### Steps to solve the equation:

1. **Combine logarithmic terms on the left-hand side:**
   Use the logarithmic property:
   \[
   \log_b{A} + \log_b{B} = \log_b{(A \cdot B)}
   \]
   So, the left side becomes:
   \[
   \log_5{(x(x + 1))} = 1 + \log_5{(x + 12)}
   \]

2. **Isolate the logarithmic terms:**
   Subtract \(\log_5{(x + 12)}\) from both sides:
   \[
   \log_5{(x(x + 1))} - \log_5{(x + 12)} = 1
   \]
   Apply the logarithmic property:
   \[
   \log_b{\frac{A}{B}} = \log_b{A} - \log_b{B}
   \]
   So the equation becomes:
   \[
   \log_5{\left(\frac{x(x + 1)}{x + 12}\right)} = 1
   \]

3. **Rewrite the equation in exponential form:**
   Recall that:
   \[
   \log_b{A} = C \quad \text{implies} \quad A = b^C
   \]
   In this case, we have:
   \[
   \frac{x(x + 1)}{x + 12} = 5^1 = 5
   \]

4. **Solve the resulting rational equation:**
   Multiply both sides by \(x + 12\) to eliminate the denominator:
   \[
   x(x + 1) = 5(x + 12)
   \]
   Expand both sides:
   \[
   x^2 + x = 5x + 60
   \]
   Rearrange the terms to set the equation to 0:
   \[
   x^2 + x - 5x - 60 = 0
   \]
   Simplify:
   \[
   x^2 - 4x - 60 = 0
   \]

5. **Solve the quadratic equation:**
   Use the quadratic formula:
   \[
   x = \frac{-(-4) \pm \sqrt{(-4)^2 - 4(1)(-60)}}{2(1)}
   \]
   Simplify:
   \[
   x = \frac{4 \pm \sqrt{16 + 240}}{2} = \frac{4 \pm \sqrt{256}}{2} = \frac{4 \pm 16}{2}
   \]
   This gives two solutions:
   \[
   x = \frac{4 + 16}{2} = \frac{20}{2} = 10
   \]
   \[
   x = \frac{4 - 16}{2} = \frac{-12}{2} = -6
   \]

6. **Check for extraneous solutions:**
   Logarithms are only defined for positive arguments, so we discard \(x = -6\) because \(\log_5{x}\) would be undefined.

   Thus, the only solution is:
   \[
   x = 10
   \]

Let me know if you need further clarification, and here are 5 related questions you can explore:

1. What is the general form of a logarithmic equation?
2. How does the product rule of logarithms help simplify logarithmic expressions?
3. Why must we discard negative solutions when solving logarithmic equations?
4. What is the relationship between logarithms and exponential functions?
5. How can we apply the change-of-base formula to evaluate logarithms?

**Tip:** Always check if the arguments inside a logarithmic function are valid (i.e., positive) before concluding the final solution.

Resuelva la siguiente ecuación log_5 x + log_5 (x + 1) = 1 + log_5 (x + 12)

Learn how to solve the logarithmic equation log_5 x + log_5 (x + 1) = 1 + log_5 (x + 12) using properties of logarithms and the quadratic formula. This problem involves logarithmic identities, exponential functions, and solving a quadratic equation, ideal for high school students.

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