Question

If \(\cosh(x-\log 3)=\sinh x\), then \(x=\)

• A. \(\dfrac{1}{2}\log 3\)
• B. \(\dfrac{1}{2}\log 6\)
• C. \(\dfrac{1}{2}\log 5\)
• D. \(\log 3\)

Hint:

For equations involving \(\sinh x\) and \(\cosh x\), first convert them into exponential form using \[ \cosh x=\frac{e^x+e^{-x}}{2} \] and \[ \sinh x=\frac{e^x-e^{-x}}{2} \]

Answer 1

The Correct Option is B

Step 1: Use hyperbolic function formulas.
We know that \[ \cosh a=\frac{e^a+e^{-a}}{2} \] and \[ \sinh x=\frac{e^x-e^{-x}}{2} \] Given, \[ \cosh(x-\log 3)=\sinh x \] So, \[ \frac{e^{x-\log 3}+e^{-(x-\log 3)}}{2} = \frac{e^x-e^{-x}}{2} \]

Step 2: Simplify the exponential terms.
Since \[ e^{x-\log 3}=\frac{e^x}{3} \] and \[ e^{-(x-\log 3)}=e^{-x+\log 3}=3e^{-x} \] Therefore, \[ \frac{\frac{e^x}{3}+3e^{-x}}{2} = \frac{e^x-e^{-x}}{2} \] Multiplying both sides by \(2\), \[ \frac{e^x}{3}+3e^{-x}=e^x-e^{-x} \]

Step 3: Solve for \(x\).
Rearranging, \[ 3e^{-x}+e^{-x}=e^x-\frac{e^x}{3} \] So, \[ 4e^{-x}=\frac{2e^x}{3} \] Multiplying by \(3\), \[ 12e^{-x}=2e^x \] Thus, \[ 6e^{-x}=e^x \] Multiplying by \(e^x\), \[ 6=e^{2x} \] Taking logarithm on both sides, \[ 2x=\log 6 \] Hence, \[ x=\frac{1}{2}\log 6 \]

Step 4: Final conclusion.
Therefore, \[ \boxed{\frac{1}{2}\log 6} \]