To find the inverse function of $f(x) = \frac{8^{2x} - 8^{-2x}}{8^{2x} + 8^{-2x}}$, we need to follow these steps:
First, let's look at the form of $f(x)$. This expression can be simplified with the help of hyperbolic functions. Recall the definitions of the hyperbolic sine and cosine:
$\sinh(u) = \frac{e^u - e^{-u}}{2}$
$\cosh(u) = \frac{e^u + e^{-u}}{2}$
We notice that $f(x)$ resembles the hyperbolic tangent function:
Recognize that $f(x) = \tanh(2 \ln(8)x)$.
To find the inverse, we set $y = f(x)$, thus $y = \tanh(2 \ln(8)x)$. The inverse of the hyperbolic tangent is the hyperbolic arctangent, so:
$x = \frac{1}{2 \ln(8)} \tanh^{-1}(y)$.
We now express the inverse in terms of base 8 logarithms:
$2 \ln 8 = 2 \cdot \log_{e} 8$
$x = \frac{1}{2 \cdot \log_{e} 8} \tanh^{-1}(y)$
Since $\tanh^{-1}(y) = \frac{1}{2} \ln \left( \frac{1+y}{1-y} \right)$, we have:
$x = \frac{1}{4 \cdot \log_{e} 8} \ln \left( \frac{1+y}{1-y} \right)$.
This simplifies to:
$x = \frac{1}{4} (\log_8 e) \ln \left( \frac{1+y}{1-y} \right)$.
Thus, the correct inverse function is $f^{-1}(x) = \frac{1}{4} (\log_8 e) \ln \left( \frac{1+x}{1-x} \right)$.
Therefore, the correct answer is the second option: $\frac{1}{4}\left(\log_{8} \,e\right) \log_{e} \left(\frac{1+x}{1-x}\right)$.
