To solve the equation:

$\log_{10} 6 + \log_{10} (x + 2) = \log_{10} (x + 5) + \log_{10} x$

We can use the properties of logarithms:

1. $\log_{10} a + \log_{10} b = \log_{10} (a \cdot b)$.

Step 1: Combine the logarithms on both sides.

• Left-hand side: $\log_{10} (6 \cdot (x + 2)) = \log_{10} (6x + 12)$
• Right-hand side: $\log_{10} ((x + 5) \cdot x) = \log_{10} (x^2 + 5x)$

Now the equation becomes: $\log_{10} (6x + 12) = \log_{10} (x^2 + 5x)$

Step 2: Remove the logarithms.

Since the logs are equal, the expressions inside must be equal: $6x + 12 = x^2 + 5x$

Step 3: Rearrange the equation.

Move all terms to one side: $x^2 + 5x - 6x - 12 = 0$ $x^2 - x - 12 = 0$

Step 4: Factor the quadratic equation.

Factor $x^2 - x - 12 = 0$: $(x - 4)(x + 3) = 0$

Step 5: Solve for $x$.

Set each factor equal to zero: $x - 4 = 0 \quad \Rightarrow \quad x = 4$ $x + 3 = 0 \quad \Rightarrow \quad x = -3$

Step 6: Check for extraneous solutions.

Logarithms are undefined for non-positive numbers, so we must check that $x + 2$, $x + 5$, and $x$ are positive.

• For $x = 4$:

  • $x + 2 = 6$, $x + 5 = 9$, and $x = 4$, all positive. So, $x = 4$ is a valid solution.

• For $x = -3$:

  • $x = -3$ is not valid because $\log_{10} (-3)$ is undefined.

Thus, the only solution is $x = 4$.

---

Would you like any further details or have any questions?

Here are five related questions to explore:

1. How do you solve more complex logarithmic equations with multiple variables?
2. What happens if there are more logarithmic terms on one side than the other?
3. How can we use logarithms to solve exponential equations?
4. What are the conditions for the existence of a logarithm?
5. How does the base of the logarithm affect the solution?

Tip: Always check the domain when solving logarithmic equations to avoid invalid solutions.