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 of the function \( f(x) = 5^{x^{(x-4)}} \), we will follow these steps:

### Step 1: Set up the equation
We start with the function:
\[ y = f(x) = 5^{x^{(x-4)}} \]
To find the inverse, we need to express \( x \) in terms of \( y \).

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

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

### Step 4: Form a quadratic equation
We can express this in a more manageable form. Let’s rewrite it as:
\[ x^{x-4} = \log_5(y) \]
This can be transformed into a quadratic equation. Let \( z = x \):
\[ z^{(z-4)} = \log_5(y) \]
This indicates that we need to solve for \( z \).

### Step 5: Solve the quadratic equation
Rearranging gives us:
\[ z^{z-4} - \log_5(y) = 0 \]
This is a transcendental equation, but we can express it as:
\[ z^2 - 4z - \log_5(y) = 0 \]
This is a quadratic equation in \( z \).

### Step 6: Use the quadratic formula
Using the quadratic formula \( z = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Here, \( a = 1 \), \( b = -4 \), and \( c = -\log_5(y) \):
\[ z = \frac{4 \pm \sqrt{(-4)^2 - 4 \cdot 1 \cdot (-\log_5(y))}}{2 \cdot 1} \]
This simplifies to:
\[ z = \frac{4 \pm \sqrt{16 + 4\log_5(y)}}{2} \]
\[ z = 2 \pm \sqrt{4 + \log_5(y)} \]

### Step 7: Determine the correct root
Since \( f(x) \) is defined for \( x \geq 4 \), we take the positive root:
\[ x = 2 + \sqrt{4 + \log_5(y)} \]

### Step 8: Replace \( y \) with \( x \)
To find the inverse function, we replace \( y \) with \( x \):
\[ f^{-1}(x) = 2 + \sqrt{4 + \log_5(x)} \]

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