Question

If \( 2\sinh x = \cosh x \), then \( x = \)

• A. \( \frac{1}{3} \log 2 \)
• B. \( 2 \log 3 \)
• C. \( \mathbf{\frac{1}{2} \log 3} \)
• D. \( \log 9 \)

Hint:

Always use the exponential definitions of hyperbolic functions to solve equations like \( \sinh x = \cosh x \) efficiently.

Answer 1

The Correct Option is C

We use the definitions: \[ \sinh x = \frac{e^x - e^{-x}}{2}, \quad \cosh x = \frac{e^x + e^{-x}}{2} \] Given: \[ 2\sinh x = \cosh x \Rightarrow 2 \cdot \frac{e^x - e^{-x}}{2} = \frac{e^x + e^{-x}}{2} \Rightarrow e^x - e^{-x} = \frac{e^x + e^{-x}}{2} \] Multiply both sides by 2: \[ 2e^x - 2e^{-x} = e^x + e^{-x} \Rightarrow (2e^x - e^x) = (2e^{-x} + e^{-x}) \Rightarrow e^x = 3e^{-x} \Rightarrow e^{2x} = 3 \Rightarrow 2x = \log 3 \Rightarrow x = \frac{1}{2} \log 3 \]