`f(x) ={{:((2^(x+2)-16)/(4^(x)-16), if x ne 2 ),(k, if x = 2):}` ,`x = 2`.

Explanation

To find the value of \( k \) such that the function \( f(x) \) is continuous at \( x = 2 \), we need to ensure that the limit of \( f(x) \) as \( x \) approaches 2 is equal to \( f(2) \).

The function is defined as follows:

\[ f(x) = \begin{cases} \frac{2^{(x+2)} - 16}{4^x - 16} & \text{if } x \neq 2 \\ k & \text{if } x = 2 \end{cases} \]

### Step 1: Calculate the limit as \( x \) approaches 2

We need to compute:

\[ \lim_{x \to 2} f(x) = \lim_{x \to 2} \frac{2^{(x+2)} - 16}{4^x - 16} \]

### Step 2: Simplify the expression

First, we notice that \( 4^x = (2^2)^x = 2^{2x} \). Thus, we can rewrite the limit as:

\[ \lim_{x \to 2} \frac{2^{(x+2)} - 16}{2^{2x} - 16} \]

Next, we can substitute \( 16 \) as \( 2^4 \):

\[ \lim_{x \to 2} \frac{2^{(x+2)} - 2^4}{2^{2x} - 2^4} \]

### Step 3: Factor the numerator and denominator

Both the numerator and denominator can be factored using the difference of squares:

\[ \text{Numerator: } 2^{(x+2)} - 2^4 = 2^{(x+2)} - 2^{(2+2)} = (2^{(x+2)} - 2^4) = (2^{(x+2)} - 2^4) = (2^{(x+2)} - 2^4)(2^{(x+2)} + 2^4) \]

\[ \text{Denominator: } 2^{2x} - 2^4 = (2^x - 4)(2^x + 4) \]

### Step 4: Substitute \( x = 2 \)

Now we can evaluate the limit:

\[ \lim_{x \to 2} \frac{(2^{(x+2)} - 16)(2^{(x+2)} + 16)}{(2^x - 4)(2^x + 4)} \]

### Step 5: Evaluate the limit

Substituting \( x = 2 \):

\[ \text{Numerator: } 2^{(2+2)} - 16 = 16 - 16 = 0 \]
\[ \text{Denominator: } 2^2 - 4 = 4 - 4 = 0 \]

Since both the numerator and denominator approach 0, we can apply L'Hôpital's Rule or simplify further.

### Step 6: Apply L'Hôpital's Rule

Differentiating the numerator and denominator:

1. Differentiate the numerator:
\[ \frac{d}{dx}(2^{(x+2)} - 16) = 2^{(x+2)} \ln(2) \]

2. Differentiate the denominator:
\[ \frac{d}{dx}(4^x - 16) = 4^x \ln(4) \]

Now we can evaluate the limit again:

\[ \lim_{x \to 2} \frac{2^{(x+2)} \ln(2)}{4^x \ln(4)} \]

Substituting \( x = 2 \):

\[ = \frac{2^{(2+2)} \ln(2)}{4^2 \ln(4)} = \frac{16 \ln(2)}{16 \ln(4)} = \frac{\ln(2)}{\ln(4)} = \frac{\ln(2)}{2\ln(2)} = \frac{1}{2} \]

### Step 7: Set the limit equal to \( k \)

Since \( f(x) \) is continuous at \( x = 2 \):

\[ k = \lim_{x \to 2} f(x) = \frac{1}{2} \]

Thus, the value of \( k \) is:

\[ \boxed{\frac{1}{2}} \]