Solve ln(a+5) + ln(a+7) = 0: Finding the Value of a

To solve this problem, we'll follow these steps:

• Step 1: Combine the logarithms using the sum rule for logarithms.
• Step 2: Solve the resulting equation for $a$.
• Step 3: Ensure the solution meets domain restrictions.

Let's work through each step:

Step 1: We have the equation $\ln(a + 5) + \ln(a + 7) = 0$. Using the property of logarithms, combine the expressions:
$\ln(a+5) + \ln(a+7) = \ln((a+5)(a+7)) = 0$.

Step 2: Knowing $\ln((a+5)(a+7)) = 0$, use the exponential property that if $\ln(x) = 0$, then $x = 1$. Thus, set the expression inside the logarithm to 1:
$(a+5)(a+7) = 1$.

Now, expand and solve the equation:
$a^2 + 12a + 35 = 1$.
Rearrange this into a quadratic form:
$a^2 + 12a + 34 = 0$.

Step 3: Solve this quadratic equation using the quadratic formula $a = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, where $a = 1, b = 12, \text{ and } c = 34$:
$a = \frac{-12 \pm \sqrt{12^2 - 4 \times 1 \times 34}}{2 \times 1}$.

Calculate the discriminant:
$b^2 - 4ac = 144 - 136 = 8$.

Insert values back into the quadratic formula:
$a = \frac{-12 \pm \sqrt{8}}{2}$.
Simplify:
$a = \frac{-12 \pm 2\sqrt{2}}{2}$ = $-6 \pm \sqrt{2}$.

Given the domain restrictions: $a+5 > 0$ and $a+7 > 0$, we calculate the solutions:
The acceptable value is $-6 + \sqrt{2}$, since the domain restriction would invalidate another potential candidate.

Therefore, the solution to the problem is $-6 + \sqrt{2}$.