`f(x) = (8^(2x) - 8^(-2x))/(8^(2x) + 8^(-2x)` find the inverse of the function

Options

A. `1/4 log_8( (1+x)/(1-x))`
B. `1/4 log_8 (1-x)/(1+x)`
C. `1/2 log_8 (1+x)/(1-x)`
D. `1/2 log_8 (1-x)/(1+x)`

Explanation

To find the inverse of the function \( f(x) = \frac{8^{2x} - 8^{-2x}}{8^{2x} + 8^{-2x}} \), we can follow these steps:

### Step 1: Set the function equal to \( y \)
Let \( f(x) = y \):
\[ y = \frac{8^{2x} - 8^{-2x}}{8^{2x} + 8^{-2x}} \]

### Step 2: Cross-multiply to eliminate the fraction
Cross-multiplying gives:
\[ y(8^{2x} + 8^{-2x}) = 8^{2x} - 8^{-2x} \]

### Step 3: Expand and rearrange the equation
Expanding the left side:
\[ y \cdot 8^{2x} + y \cdot 8^{-2x} = 8^{2x} - 8^{-2x} \]
Rearranging gives:
\[ y \cdot 8^{2x} - 8^{2x} = -y \cdot 8^{-2x} - 8^{-2x} \]
Factoring out \( 8^{2x} \) on the left:
\[ ( y - 1 ) 8^{2x} = -( y + 1 ) 8^{-2x} \]

### Step 4: Multiply both sides by \( 8^{2x} \)
This gives:
\[ ( y - 1 ) ( 8^{2x} )^2 = -( y + 1 ) \]
Let \( z = 8^{2x} \), then we have:
\[ ( y - 1 ) z^2 = -( y + 1 ) \]

### Step 5: Solve for \( z^2 \)
Rearranging gives:
\[ z^2 = \frac{-(y + 1)}{y - 1} \]

### Step 6: Take the square root
Taking the square root gives:
\[ z = \sqrt{\frac{-(y + 1)}{y - 1}} \]
Since \( z = 8^{2x} \), we can express it as:
\[ 8^{2x} = \sqrt{\frac{-(y + 1)}{y - 1}} \]

### Step 7: Take logarithm to solve for \( x \)
Taking logarithm base 8:
\[ 2x = \log_8 \left( \sqrt{\frac{-(y + 1)}{y - 1}} \right) \]
Thus,
\[ x = \frac{1}{2} \log_8 \left( \sqrt{\frac{-(y + 1)}{y - 1}} \right) \]

### Step 8: Simplify the expression
Using properties of logarithms:
\[ x = \frac{1}{4} \log_8 \left( \frac{-(y + 1)}{y - 1} \right) \]

### Step 9: Write the inverse function
Thus, the inverse function \( f^{-1}(x) \) is:
\[ f^{-1}(x) = \frac{1}{4} \log_8 \left( \frac{-(x + 1)}{x - 1} \right) \]

### Summary
The inverse of the function \( f(x) = \frac{8^{2x} - 8^{-2x}}{8^{2x} + 8^{-2x}} \) is:
\[ f^{-1}(x) = \frac{1}{4} \log_8 \left( \frac{-(x + 1)}{x - 1} \right) \]