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

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Question

(D) Given the function $y = \frac{10^x - 10^{-x}}{10^x + 10^{-x}} + 1$. Subtract $1$ from both sides: $y - 1 = \frac{10^x - 10^{-x}}{10^x + 10^{-x}}$.

Vedclass · Accepted Answer

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

Vedclass · Answer

(D) Given the function $y = \frac{10^x - 10^{-x}}{10^x + 10^{-x}} + 1$. Subtract $1$ from both sides: $y - 1 = \frac{10^x - 10^{-x}}{10^x + 10^{-x}}$. Let $u = 10^x$. Then $10^{-x} = \frac{1}{u}$. So, $y - 1 = \frac{u - 1/u}{u + 1/u} = \frac{u^2 - 1}{u^2 + 1}$. Let $Y = y - 1$. Then $Y(u^2 + 1) = u^2 - 1$. $Yu^2 + Y = u^2 - 1 \implies Y + 1 = u^2(1 - Y)$. $u^2 = \frac{1 + Y}{1 - Y} = \frac{1 + (y - 1)}{1 - (y - 1)} = \frac{y}{2 - y}$. Since $u = 10^x$, we have $10^{2x} = \frac{y}{2 - y}$. Taking $\log_{10}$ on both sides: $2x = \log_{10} \left(\frac{y}{2 - y}\right)$. Therefore, $x = \frac{1}{2} \log_{10} \left(\frac{y}{2 - y}\right)$.

The inverse of the function $y = \frac{10^x - 10^{-x}}{10^x + 10^{-x}} + 1$ is $x =$

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