The inverse of the function y = frac{10^x - 1 | vedclass

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(A) Given the function $y = \frac{10^x - 10^{-x}}{10^x + 10^{-x}}$.Multiply the numerator and denominator by $10^x$:$y = \frac{10^{2x} - 1}{10^{2x} +

Vedclass · Accepted Answer

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

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(A) Given the function $y = \frac{10^x - 10^{-x}}{10^x + 10^{-x}}$.Multiply the numerator and denominator by $10^x$:$y = \frac{10^{2x} - 1}{10^{2x} + 1}$Now,solve for $x$ in terms of $y$:$y(10^{2x} + 1) = 10^{2x} - 1$$y \cdot 10^{2x} + y = 10^{2x} - 1$$1 + y = 10^{2x} - y \cdot 10^{2x}$$1 + y = 10^{2x}(1 - y)$$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 find the inverse function $f^{-1}(x)$:$f^{-1}(x) = \frac{1}{2} \log_{10} \left( \frac{1 + x}{1 - x} \right)$

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

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