Question

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

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

Hint:

To find an inverse function \( f^{-1}(x) \), set \( y = f(x) \), then algebraically solve for \( x \) in terms of \( y \). Finally, swap the variables \( x \) and \( y \). Using the componendo and dividendo rule on \( y/1 = (10^{2x}-1)/(10^{2x}+1) \) can also quickly yield \( (1+y)/(1-y) = 10^{2x} \).

Answer 1

The Correct Option is B

To find the inverse of the function, let \( y = f(x) \). \[ y = \frac{10^x - 10^{-x}}{10^x + 10^{-x}} \]
Multiply the numerator and denominator by \( 10^x \): \[ y = \frac{10^{2x} - 1}{10^{2x} + 1} \]
Now, we solve for \( x \) in terms of \( y \). \[ y(10^{2x} + 1) = 10^{2x} - 1 \] \[ y \cdot 10^{2x} + y = 10^{2x} - 1 \] \[ y + 1 = 10^{2x} - y \cdot 10^{2x} \] \[ y + 1 = 10^{2x}(1 - y) \] \[ 10^{2x} = \frac{1+y}{1-y} \]
Take \( \log_{10} \) on both sides: \[ 2x = \log_{10}\left(\frac{1+y}{1-y}\right) \] \[ x = \frac{1}{2} \log_{10}\left(\frac{1+y}{1-y}\right) \]
Now, replace \( x \) with \( f^{-1}(x) \) and \( y \) with \( x \): \[ f^{-1}(x) = \frac{1}{2} \log_{10}\left(\frac{1+x}{1-x}\right) \]