The given function is:

$f(x) = 2^{12x} - 5$

To find the range, let's analyze it step by step:

1. Base Function: The function $g(x) = 2^{12x}$ is an exponential function, and exponential functions always have positive outputs, with $g(x)$ taking values in the interval $(0, \infty)$.

2. Subtracting 5: The transformation $f(x) = 2^{12x} - 5$ shifts the graph of $g(x) = 2^{12x}$ downward by 5 units. This means that instead of ranging from $(0, \infty)$, the function now ranges from $(-5, \infty)$.

Thus, the range of the function $f(x) = 2^{12x} - 5$ is:

$\boxed{(-5, \infty)}$

Let me know if you'd like further details or have any questions.

Here are 5 related questions to explore:

1. What is the domain of $f(x) = 2^{12x} - 5$?
2. How does changing the exponent $12x$ affect the graph of $f(x)$?
3. What is the horizontal asymptote of the function $f(x) = 2^{12x} - 5$?
4. How does the function $f(x) = 2^{kx} - 5$ behave for different values of $k$?
5. What would happen to the range if the function were $f(x) = 2^{12x} + 5$