Question

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

• A. $\log_{10} (2 - x)$
• B. $\frac{1}{2} \log 10 (2x -1)$
• C. $\frac{1}{4} \log_{10} \frac{2x}{2 -x}$
• D. $\frac{1}{2} \log_{10} \frac{ 1 + x}{1 -x}$

Answer 1

Answer: (D)

$\frac{1}{2} \log_{10} \frac{ 1 + x}{1 -x}$

Answer 2

The given function is : $y = \frac{10^{x} - 10^{-x}}{10^x + 10^{-x}}$ By componendo and dividendo, we get, $\frac{1+y}{1-y} = \frac{2.10^{x}}{2.10^{-x}}$ $\Rightarrow \frac{1+y}{1-y} = 10^{2x}$ Taking $\log_{10}$ on both sides, we get, $\Rightarrow 2x = \log_{10} \left(\frac{1+y}{1-y} \right) \Rightarrow x = \frac{1}{2} \log_{10} \left(\frac{1+y}{1-y}\right)$ $\therefore f^{-1} \left(y\right) = \frac{1}{2} \log_{10} \left( \frac{1+y}{1-y}\right)$ Now replacing y with x, we get inverse, $\therefore f^{-1} \left(x\right) = \frac{1}{2} \log_{10} \left(\frac{1+x}{1-x}\right)$