Let `f:[4,oo)to[4,oo)` be defined by `f(x)=5^(x^((x-4)))`.Then `f^(-1)(x)` is

Options

A. `2-sqrt(4-logs x)`
B. `2+sqrt(4+logs x)`
C. `((1)/5)^(x^(x+4))`
D. not defined

Explanation

To find the inverse function \( f^{-1}(x) \) for the function \( f(x) = 5^{x^{(x-4)}} \), we will follow these steps:

### Step 1: Set up the equation
Let \( y = f(x) \). Thus, we have:
\[ y = 5^{x^{(x-4)}} \]

### Step 2: Take the logarithm
Taking the logarithm (base 5) of both sides gives:
\[ \log_5(y) = x^{(x-4)} \]

### Step 3: Rearranging the equation
We can rearrange the equation to express it as:
\[ x^{(x-4)} = \log_5(y) \]

### Step 4: Rewrite in terms of \( x \)
We need to solve for \( x \). To do this, we will rewrite the equation:
\[ x^{(x-4)} - \log_5(y) = 0 \]

### Step 5: Use the quadratic formula
To solve for \( x \), we can express \( x^{(x-4)} \) in a different form. Let’s denote \( z = x \) and rewrite the equation:
\[ z^{(z-4)} = \log_5(y) \]
This can be transformed into a quadratic equation in terms of \( z \):
\[ z^2 - 4z - \log_5(y) = 0 \]

### Step 6: Apply the quadratic formula
Using the quadratic formula \( z = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Here, \( a = 1 \), \( b = -4 \), and \( c = -\log_5(y) \):
\[ z = \frac{4 \pm \sqrt{(-4)^2 - 4 \cdot 1 \cdot (-\log_5(y))}}{2 \cdot 1} \]

### Step 7: Simplify the expression
Calculating the discriminant:
\[ z = \frac{4 \pm \sqrt{16 + 4\log_5(y)}}{2} \]
This simplifies to:
\[ z = 2 \pm \sqrt{4 + \log_5(y)} \]

### Step 8: Choose the positive root
Since \( f(x) \) is defined for \( x \geq 4 \) and the output must also be valid in the range, we choose the positive root:
\[ x = 2 + \sqrt{4 + \log_5(y)} \]

### Step 9: Substitute back for \( f^{-1}(x) \)
Substituting back \( y \) with \( x \) gives:
\[ f^{-1}(x) = 2 + \sqrt{4 + \log_5(x)} \]

### Final Answer
Thus, the inverse function is:
\[ f^{-1}(x) = 2 + \sqrt{4 + \log_5(x)} \]