We are tasked with solving the equation:

$4 \log(x + 5) + 4 \log(x + 2) = 0$

Step 1: Simplify the equation

We can factor out the 4 from both terms on the left-hand side:

$4 (\log(x + 5) + \log(x + 2)) = 0$

Divide both sides by 4:

$\log(x + 5) + \log(x + 2) = 0$

Step 2: Apply logarithmic properties

Using the property of logarithms, $\log(a) + \log(b) = \log(ab)$, we can combine the logarithms:

$\log((x + 5)(x + 2)) = 0$

Step 3: Exponentiate both sides

Exponentiate both sides of the equation to remove the logarithm. Recall that $\log_b(a) = c$ implies $b^c = a$. Since we are dealing with a natural logarithm (logarithm base 10), this gives us:

$(x + 5)(x + 2) = 10^0 = 1$

Step 4: Expand the equation

Now, expand the left-hand side:

$(x + 5)(x + 2) = x^2 + 2x + 5x + 10 = x^2 + 7x + 10$

Set this equal to 1:

$x^2 + 7x + 10 = 1$

Step 5: Solve the quadratic equation

Subtract 1 from both sides:

$x^2 + 7x + 9 = 0$

We now solve this quadratic equation using the quadratic formula:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

For $a = 1$, $b = 7$, and $c = 9$:

$x = \frac{-7 \pm \sqrt{7^2 - 4(1)(9)}}{2(1)}$ $x = \frac{-7 \pm \sqrt{49 - 36}}{2}$ $x = \frac{-7 \pm \sqrt{13}}{2}$

Thus, the solutions are:

$x = \frac{-7 + \sqrt{13}}{2} \quad \text{or} \quad x = \frac{-7 - \sqrt{13}}{2}$

Step 6: Check for domain restrictions

Logarithmic functions are only defined for positive arguments. Therefore, both $x + 5 > 0$ and $x + 2 > 0$ must hold. This implies:

$x > -5 \quad \text{and} \quad x > -2$

Therefore, the solutions must satisfy $x > -2$.

Conclusion:

The solution $\frac{-7 - \sqrt{13}}{2}$ is less than -2 and thus invalid. The only valid solution is:

$x = \frac{-7 + \sqrt{13}}{2}$

This is the only solution to the given equation.

Would you like further details or have any questions?

Here are five related questions you might find useful:

1. How do we determine when logarithmic arguments are valid?
2. What is the importance of domain restrictions in logarithmic equations?
3. How can we simplify logarithmic expressions further using properties of logs?
4. How does the quadratic formula help in solving equations like these?
5. Can we graphically interpret the solutions of this equation?

Tip: Always check the domain of the function before accepting all the solutions from a quadratic equation involving logarithms.