Animated Solution for Mathematics - Functions: The inverse function of $f(x)=\frac{8^{2x}-8^{-2x}}{8^{2x}+8^{-2x}}$, $x\in (-1,1)$, is

Welcome, future engineer! Today, we are going to embark on a journey to demystify the inverse of a function. When you first look at $f(x)=\frac{8^{2x}-8^{-2x}}{8^{2x}+8^{-2x}}$, it might seem intimidating.

It looks like a complex exponential beast, but I want you to see it for what it truly is: a beautiful, symmetric structure waiting to be unlocked. In the world of JEE Advanced, we don't just solve problems; we perform algebraic surgery.

Our goal is to find the inverse, which geometrically means reflecting the function across the line $y=x$. To do this, we must isolate $x$.

The Algebraic Cleanup

The first thing that catches our eye is the presence of negative exponents, $8^{-2x}$. They are the clutter in our workspace. To clear them, we perform a simple yet powerful operation: multiply both the numerator and the denominator by $8^{2x}$.

When we distribute this, the expression transforms. The numerator becomes $8^{2x}\cdot 8^{2x}-8^{-2x}\cdot 8^{2x}$, which simplifies to $8^{4x}-1$. Similarly, the denominator becomes $8^{4x}+1$.

Suddenly, the function looks much cleaner:

We have tamed the beast.

The Elegant Shortcut

Now, we set $y=\frac{8^{4x}-1}{8^{4x}+1}$. We need to isolate $x$. You could cross-multiply, but I want you to think like a topper.

When you see a fraction equal to a variable, think of Componendo and Dividendo. This theorem states that if $\frac{a}{b}=\frac{c}{d}$, then $\frac{a+b}{a-b}=\frac{c+d}{c-d}$.

Applying this to our equation $\frac{y}{1}=\frac{8^{4x}-1}{8^{4x}+1}$, we get:

Look at how the terms cancel out! The numerator becomes $2\cdot 8^{4x}$ and the denominator becomes $-2$. The twos vanish, leaving us with:

This is the elegance of mathematics.

The Final Extraction

We are almost there. We have $\frac{y+1}{y-1}=-8^{4x}$. Let's multiply by $-1$ to clean up that negative sign, giving us:

Now, how do we get $x$ out of the exponent? We use the logarithm. Taking ${\log }_8$ on both sides, we get:

Dividing by $4$, we find $x=\frac{1}{4}{\log }_8\left(\frac{1+y}{1-y}\right)$. Finally, we look at our options. They are in base $e$.

We apply the base change formula: ${\log }_{a}b={\log }_{a}e\cdot \ln b$. Thus, our inverse function is:

You have successfully navigated the complexity and arrived at the solution. Remember, every complex problem is just a series of simple steps waiting for you to take them. Keep practicing, and keep that curiosity alive!