Problem 181

To solve the problem, we need to understand the definition of sets $M$ and $N$ and find their intersection after determining the complement of set $N$ with respect to the real numbers.

Firstly, the set $M$ is defined as all $x$ that are greater than or equal to $-2$. In interval notation, we have:
$$M = [ -2, +\infty )$$

For set $N$, we need to find all $x$ that satisfy $2^x - 1 > 0$. Solving for $x$ yields:
$$2^x > 1$$
This inequality is true for all $x > 0$, as any positive exponent of 2 results in a number greater than 1.

Therefore, we have:
$$N = ( 0, +\infty )$$

Now, the complement of $N$ in the real numbers, $C_{\mathbb{R}}N$), consists of all real numbers that are not in $N$. Hence, it includes all real numbers less than or equal to $0$. Given this, we can express $C_{\mathbb{R}}N$ as:
$$C_{\mathbb{R}}N = ( -\infty, 0 ]$$

Now we need to find $M \cap C_{\mathbb{R}}N$, the intersection of sets $M$ and $C_{\mathbb{R}}N$. This will be all numbers that are in both sets. As $M$ includes numbers from $-2$ upward, and $C_{\mathbb{R}}N$ includes numbers up to $0$, their intersection will include all numbers from $-2$ up to $0$, inclusive.

Therefore:
$$M \cap C_{\mathbb{R}}N = [-2, 0]$$

Hence, the correct answer is:

\boxed{D: \left\{ x \mid -2 \leq x \leq 0 \right\}}