Question:
The value of
\[
\tanh(\log x)
\]
is
The value of
\[
\tanh(\log x)
\]
is
Show Hint
For expressions involving \(\log x\) inside hyperbolic functions, substitute \(e^{\log x}=x\) and then simplify using the definitions of hyperbolic functions.
Updated On: Jun 26, 2026
- \(\dfrac{x+1}{x-1}\)
- \(\dfrac{x^2+1}{x^2-1}\)
- \(\dfrac{x^2-1}{x^2+1}\)
- \(2x\)
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The Correct Option is C
Solution and Explanation
Step 1: Use the definition of hyperbolic tangent.
We know that \[ \tanh t=\frac{e^t-e^{-t}}{e^t+e^{-t}} \] Let \[ t=\log x \] Then \[ \tanh(\log x) = \frac{e^{\log x}-e^{-\log x}} {e^{\log x}+e^{-\log x}} \]
Step 2: Simplify the exponential terms.
Since \[ e^{\log x}=x \] and \[ e^{-\log x}=\frac{1}{x}, \] we get \[ \tanh(\log x) = \frac{x-\frac1x}{x+\frac1x} \]
Step 3: Remove the fractions.
Multiplying numerator and denominator by \(x\), \[ \tanh(\log x) = \frac{x^2-1}{x^2+1} \]
Step 4: Final conclusion.
Therefore, \[ \boxed{\tanh(\log x)=\frac{x^2-1}{x^2+1}} \] Hence, the correct option is \[ \boxed{(3)} \]
We know that \[ \tanh t=\frac{e^t-e^{-t}}{e^t+e^{-t}} \] Let \[ t=\log x \] Then \[ \tanh(\log x) = \frac{e^{\log x}-e^{-\log x}} {e^{\log x}+e^{-\log x}} \]
Step 2: Simplify the exponential terms.
Since \[ e^{\log x}=x \] and \[ e^{-\log x}=\frac{1}{x}, \] we get \[ \tanh(\log x) = \frac{x-\frac1x}{x+\frac1x} \]
Step 3: Remove the fractions.
Multiplying numerator and denominator by \(x\), \[ \tanh(\log x) = \frac{x^2-1}{x^2+1} \]
Step 4: Final conclusion.
Therefore, \[ \boxed{\tanh(\log x)=\frac{x^2-1}{x^2+1}} \] Hence, the correct option is \[ \boxed{(3)} \]
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