We are tasked with finding the approximate zero of the function $f(x) = \ln(x^2 - 4) - 8$. To do this, we need to solve for $x$ when $f(x) = 0$.

So, set the equation equal to zero:

$\ln(x^2 - 4) - 8 = 0$

Rearranging this gives:

$\ln(x^2 - 4) = 8$

Exponentiating both sides to eliminate the natural logarithm:

$x^2 - 4 = e^8$

Now solve for $x^2$:

$x^2 = e^8 + 4$

Taking the square root:

$x = \pm \sqrt{e^8 + 4}$

We can now approximate this expression. Using $e^8 \approx 2980.96$:

$x = \pm \sqrt{2980.96 + 4} \approx \pm \sqrt{2984.96} \approx \pm 54.63$

So the approximate zero is $x \approx 55$.

Thus, the correct answer is:

A: $x \approx 55$.

Would you like further details on the steps?

Here are five related questions to expand this topic:

1. How do you solve logarithmic equations when other constants are involved?
2. What is the significance of taking the natural logarithm in equations?
3. How can logarithmic and exponential functions be applied to real-world problems?
4. What are the domain restrictions for logarithmic functions like $\ln(x^2 - 4)$?
5. How do different bases of logarithms affect the solutions of equations?

Tip: Always check the domain of the function before solving logarithmic equations, as the logarithm function requires positive arguments.