For the following two (02) items: Consider the function f(x) = 10^x...

Let $y = f(x) = \dfrac{10^x - 10^{-x}}{10^x + 10^{-x}}$

Multiplying the numerator and the denominator by $10^x$, we get:

$y = \dfrac{10^{2x} - 1}{10^{2x} + 1}$

Applying componendo and dividendo:

$\dfrac{1 + y}{1 - y} = \dfrac{10^{2x} + 1 + 10^{2x} - 1}{10^{2x} + 1 - (10^{2x} - 1)}$

$\dfrac{1 + y}{1 - y} = \dfrac{2 \cdot 10^{2x}}{2}$

$10^{2x} = \dfrac{1 + y}{1 - y}$

Taking logarithm to the base $10$ on both sides:

$2x = \log_{10}\left(\dfrac{1 + y}{1 - y}\right)$

$x = \dfrac{1}{2}\log_{10}\left(\dfrac{1 + y}{1 - y}\right)$

Replacing $y$ with $x$ to find the inverse function:

$f^{-1}(x) = \dfrac{1}{2}\log_{10}\left(\dfrac{1 + x}{1 - x}\right)$

Answer: $\dfrac{1}{2}\log_{10}\left(\dfrac{1 + x}{1 - x}\right)