The inverse of the function f(x) = frac{10^x | vedclass

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Question

(A) Let $y = \frac{10^x - 10^{-x}}{10^x + 10^{-x}}$.To find the inverse,we solve for $x$ in terms of $y$.Multiply the numerator and denominator by $10

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$\frac{1}{2} \log_{10} \left( \frac{1+x}{1-x} \right)$

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(A) Let $y = \frac{10^x - 10^{-x}}{10^x + 10^{-x}}$.To find the inverse,we solve for $x$ in terms of $y$.Multiply the numerator and denominator by $10^x$:$y = \frac{10^{2x} - 1}{10^{2x} + 1}$.$y(10^{2x} + 1) = 10^{2x} - 1$.$y \cdot 10^{2x} + y = 10^{2x} - 1$.$1 + y = 10^{2x}(1 - y)$.$10^{2x} = \frac{1+y}{1-y}$.Taking $\log_{10}$ on both sides:$2x = \log_{10} \left( \frac{1+y}{1-y} \right)$.$x = \frac{1}{2} \log_{10} \left( \frac{1+y}{1-y} \right)$.Replacing $y$ with $x$ to get the inverse function:$f^{-1}(x) = \frac{1}{2} \log_{10} \left( \frac{1+x}{1-x} \right)$.

The inverse of the function $f(x) = \frac{10^x - 10^{-x}}{10^x + 10^{-x}}$ is

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