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

Options

A. `(1)/(4)(log_(g)e)log_(e)((1+x)/(1-x))`
B. `(1)/(4)(log_(g)e)log_(e)((1-x)/(1+x))`
C. `(1)/(4)log_(e)((1-x)/(1+x))`
D. `(1)/(4)log_(8)((1+x)/(1-x))`

Answer

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

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

### Step 2: Apply the component and dividendo rule
Using the component and dividendo rule, we can rewrite the equation:
\[ \frac{a - b}{a + b} = \frac{c - d}{c + d} \implies \frac{y + 1}{y - 1} = \frac{8^{2x} - 8^{-2x} + 8^{2x} + 8^{-2x}}{8^{2x} - 8^{-2x} - (8^{2x} + 8^{-2x})} \]
This simplifies to:
\[ \frac{y + 1}{y - 1} = \frac{2 \cdot 8^{2x}}{-2 \cdot 8^{-2x}} = -\frac{8^{2x}}{8^{-2x}} = -8^{4x} \]

### Step 3: Rearrange the equation
Rearranging gives us:
\[ \frac{y + 1}{1 - y} = 8^{4x} \]

### Step 4: Take logarithm on both sides
Taking the logarithm (base 8) of both sides:
\[ \log_8\left(\frac{y + 1}{1 - y}\right) = 4x \]

### Step 5: Solve for \( x \)
Now, we can solve for \( x \):
\[ x = \frac{1}{4} \log_8\left(\frac{y + 1}{1 - y}\right) \]

### Step 6: Express \( f^{-1}(x) \)
To express the inverse function, we replace \( y \) with \( x \):
\[ f^{-1}(x) = \frac{1}{4} \log_8\left(\frac{x + 1}{1 - x}\right) \]

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