Find the value of k so that the function `f(x) = {((2^(x+2) - 16)/(4^(x) - 16),"if",x ne 2),(" k","if",x = 2):}` is continuous at x = 2.

Answer

To find the value of \( k \) such that the function

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

is continuous at \( x = 2 \), we need to ensure that

\[ \lim_{x \to 2} f(x) = f(2). \]

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

We need to evaluate

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

### Step 2: Simplify the expression.

First, we can rewrite \( 4^x \) as \( (2^2)^x = 2^{2x} \). Thus, we have:

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

### Step 3: Substitute \( x = 2 \) directly.

Substituting \( x = 2 \):

- The numerator becomes \( 2^{(2+2)} - 16 = 2^4 - 16 = 16 - 16 = 0 \).
- The denominator becomes \( 2^{(2 \cdot 2)} - 16 = 2^4 - 16 = 16 - 16 = 0 \).

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

### Step 4: Factor the expression.

We can factor the numerator and denominator.

The numerator can be factored as:

\[ 2^{(x+2)} - 16 = 2^{(x+2)} - 2^4 = (2^{(x+2)} - 2^4) = (2^{(x+2)} - 2^4) = (2^{(x+2)} - 2^4). \]

The denominator can be factored as:

\[ 4^x - 16 = 2^{2x} - 16 = (2^x - 4)(2^x + 4). \]

### Step 5: Rewrite the limit.

Thus, we can rewrite the limit as:

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

### Step 6: Cancel common factors.

Now, we can cancel \( 2^{(x+2)} - 16 \) with \( 2^x - 4 \) after factoring both:

\[ \lim_{x \to 2} \frac{4}{2^x + 4}. \]

### Step 7: Substitute \( x = 2 \) again.

Now substituting \( x = 2 \):

\[ \lim_{x \to 2} \frac{4}{2^2 + 4} = \frac{4}{4 + 4} = \frac{4}{8} = \frac{1}{2}. \]

### Step 8: Set the limit equal to \( k \).

Since \( f(2) = k \) and for continuity we need:

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

### Conclusion

Thus, the value of \( k \) that makes the function continuous at \( x = 2 \) is

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