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: Write out cosh and sinh in exponential form.
\(\cosh(x-\log3)=\frac{e^{x-\log3}+e^{-(x-\log3)}}{2}=\frac{e^x/3+3e^{-x}}{2}\). And \(\sinh x=\frac{e^x-e^{-x}}{2}\).

Step 2: Set equal and solve.
\(\frac{e^x}{3}+3e^{-x}=e^x-e^{-x}\). Multiply by \(3e^x\): \(e^{2x}+9=3e^{2x}-3\Rightarrow 2e^{2x}=12\Rightarrow e^{2x}=6\Rightarrow x=\frac{1}{2}\log6\). \[ \boxed{\dfrac{1}{2}\log 6} \]