If `f(x) ={{:((2^(x+2)-16)/(4^(x)-16), if x ne 2 ),(k, if x = 2):}` ,` is 'conti nuous' at x = 2`. then find 'k'.

Explanation

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

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

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

We start by evaluating the limit:

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

### Step 2: Simplify the expression

First, we can rewrite \( 2^{(x+2)} \) as \( 4 \cdot 2^x \):

\[ \lim_{x \to 2} \frac{4 \cdot 2^x - 16}{4^x - 16}. \]

### Step 3: Factor the numerator and denominator

Next, we can factor both the numerator and the denominator.

- The numerator can be factored as:

\[ 4 \cdot 2^x - 16 = 4(2^x - 4). \]

- The denominator can be rewritten using \( 4^x = (2^2)^x = 2^{2x} \):

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

### Step 4: Substitute the factored forms back into the limit

Now, substituting back, we have:

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

### Step 5: Cancel common factors

Since \( 2^x - 4 \) is a common factor in both the numerator and the denominator, we can cancel it (as long as \( x \neq 2 \)):

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

### Step 6: Evaluate the limit

Now we can directly substitute \( x = 2 \):

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

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

Since we want the function to be continuous at \( x = 2 \), we set:

\[ k = \frac{1}{2}. \]

### Final Answer

Thus, the value of \( k \) is

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