Question

\(\tanh(\log x) = \) (Assuming \(\log x\) is natural logarithm \(\ln x\))

• A. \( \frac{x+1}{x-1} \)
• B. \( \frac{x^2+1}{x^2-1} \)
• C. \( \frac{x^2-1}{x^2+1} \)
• D. \( 2x \)

Hint:

Definition of hyperbolic tangent: \(\tanh y = \frac{e^y - e^{-y}}{e^y + e^{-y}}\).
If \(\log x\) means natural logarithm \(\ln x\), then \(e^{\ln x} = x\) and \(e^{-\ln x} = 1/x\).
Substitute and simplify the algebraic fraction.

Answer 1

The Correct Option is C

The hyperbolic tangent function is defined as: \[ \tanh y = \frac{\sinh y}{\cosh y} = \frac{e^y - e^{-y}}{e^y + e^{-y}} \] Let \(y = \log x = \ln x\). (Assuming natural logarithm, which is standard in such contexts). Then \(e^y = e^{\ln x} = x\). And \(e^{-y} = e^{-\ln x} = e^{\ln(x^{-1})} = x^{-1} = \frac{1}{x}\). Substitute these into the definition of \(\tanh y\): \[ \tanh(\ln x) = \frac{x - \frac{1}{x}}{x + \frac{1}{x}} \] To simplify, multiply the numerator and denominator by \(x\): \[ \tanh(\ln x) = \frac{x(x - \frac{1}{x})}{x(x + \frac{1}{x})} = \frac{x^2 - 1}{x^2 + 1} \] This matches option (c). \[ \boxed{\frac{x^2-1}{x^2+1}} \]