Suppose `f:[1,infty) to [1, infty)` is defined by `f(x) = 1/2(1+ sqrt(1+4 log_(2)x))`, then `f^(-1)(3)` = …………………

Answer

To find \( f^{-1}(3) \) for the function defined as \( f(x) = \frac{1}{2} (1 + \sqrt{1 + 4 \log_2 x}) \), we will follow these steps:

### Step 1: Set up the equation for the inverse
We start by setting \( y = f(x) \):
\[ y = \frac{1}{2} (1 + \sqrt{1 + 4 \log_2 x}) \]

### Step 2: Solve for \( x \) in terms of \( y \)
Multiply both sides by 2:
\[ 2y = 1 + \sqrt{1 + 4 \log_2 x} \]
Subtract 1 from both sides:
\[ 2y - 1 = \sqrt{1 + 4 \log_2 x} \]

### Step 3: Square both sides
Now, square both sides to eliminate the square root:
\[ (2y - 1)^2 = 1 + 4 \log_2 x \]

### Step 4: Expand the left side
Expanding the left side gives:
\[ 4y^2 - 4y + 1 = 1 + 4 \log_2 x \]
Subtract 1 from both sides:
\[ 4y^2 - 4y = 4 \log_2 x \]

### Step 5: Divide by 4
Now, divide everything by 4:
\[ y^2 - y = \log_2 x \]

### Step 6: Exponentiate to solve for \( x \)
To solve for \( x \), we exponentiate both sides:
\[ x = 2^{y^2 - y} \]

### Step 7: Write the inverse function
Thus, the inverse function is:
\[ f^{-1}(y) = 2^{y^2 - y} \]

### Step 8: Find \( f^{-1}(3) \)
Now we substitute \( y = 3 \):
\[ f^{-1}(3) = 2^{3^2 - 3} \]
Calculating the exponent:
\[ 3^2 - 3 = 9 - 3 = 6 \]
So,
\[ f^{-1}(3) = 2^6 = 64 \]

### Final Answer
Thus, \( f^{-1}(3) = 64 \).

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