Question

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

• A. \( \log\left(\frac{y}{2-y}\right) \)
• B. \( \log_{10}\left(\frac{y}{2-y}\right) \)
• C. \( \frac{1}{10}\log\left(\frac{y}{1-y}\right) \)
• D. \( \frac{1}{2}\log_{10}\left(\frac{y}{2-y}\right) \)

Hint:

When dealing with expressions like \( \frac{e^x - e^{-x}}{e^x + e^{-x}} \), recognizing it as \( \tanh(x) \) or multiplying by \( e^x \) (in this case \( 10^x \)) usually simplifies the algebra significantly. Also, componendo and dividendo can be a shortcut here: if \( \frac{A}{B} = \frac{C}{D} \), then \( \frac{A+B}{A-B} = \frac{C+D}{C-D} \).

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:

To find the inverse of a function \( y = f(x) \), we need to express \( x \) in terms of \( y \).
Step 2: Detailed Explanation:

Given the function: \[ y = \frac{10^x - 10^{-x}}{10^x + 10^{-x}} + 1 \] Subtract 1 from both sides: \[ y - 1 = \frac{10^x - 10^{-x}}{10^x + 10^{-x}} \] Multiply numerator and denominator of the fraction by \( 10^x \) to simplify: \[ y - 1 = \frac{10^x(10^x - 10^{-x})}{10^x(10^x + 10^{-x})} = \frac{10^{2x} - 1}{10^{2x} + 1} \] Let \( u = 10^{2x} \). Then: \[ y - 1 = \frac{u - 1}{u + 1} \] Now, solve for \( u \) in terms of \( y \): \[ (y - 1)(u + 1) = u - 1 \] \[ u(y - 1) + (y - 1) = u - 1 \] \[ u(y - 1) - u = -1 - (y - 1) \] \[ u(y - 1 - 1) = -1 - y + 1 \] \[ u(y - 2) = -y \] \[ u = \frac{-y}{y - 2} = \frac{y}{2 - y} \] Now substitute back \( u = 10^{2x} \): \[ 10^{2x} = \frac{y}{2 - y} \] Take \( \log_{10} \) on both sides: \[ 2x = \log_{10}\left(\frac{y}{2 - y}\right) \] \[ x = \frac{1}{2} \log_{10}\left(\frac{y}{2 - y}\right) \]
Step 3: Final Answer:

The inverse function is \( x = \frac{1}{2} \log_{10}\left(\frac{y}{2 - y}\right) \).