Question

Let \(\theta\in \mathbb{R}\) such that \[ 3\sinh(2\theta)=13-3e^{2\theta}, \] then \(\theta=\)

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

Hint:

For equations involving \(\sinh x\), use \[ \sinh x=\frac{e^x-e^{-x}}{2} \] and then substitute \(e^x=t\) to convert the equation into an algebraic equation.

Answer 1

The Correct Option is A

Step 1: Use the definition of hyperbolic sine.
We know that \[ \sinh x=\frac{e^x-e^{-x}}{2} \] Therefore, \[ \sinh(2\theta)=\frac{e^{2\theta}-e^{-2\theta}}{2} \]

Step 2: Substitute in the given equation.
Given, \[ 3\sinh(2\theta)=13-3e^{2\theta} \] So, \[ 3\left(\frac{e^{2\theta}-e^{-2\theta}}{2}\right)=13-3e^{2\theta} \]

Step 3: Put \(e^{2\theta}=t\).
Let \[ e^{2\theta}=t \] Then, \[ e^{-2\theta}=\frac{1}{t} \] So, \[ \frac{3}{2}\left(t-\frac{1}{t}\right)=13-3t \]

Step 4: Remove the denominator.
Multiplying both sides by \(2t\), we get \[ 3(t^2-1)=26t-6t^2 \] \[ 3t^2-3=26t-6t^2 \] \[ 9t^2-26t-3=0 \]

Step 5: Solve the quadratic equation.
\[ 9t^2-26t-3=0 \] Using factorization, \[ 9t^2-27t+t-3=0 \] \[ 9t(t-3)+1(t-3)=0 \] \[ (t-3)(9t+1)=0 \] So, \[ t=3 \quad \text{or} \quad t=-\frac{1}{9} \]

Step 6: Use \(t=e^{2\theta}\gt 0\).
Since \[ t=e^{2\theta}\gt 0, \] we reject \[ t=-\frac{1}{9} \] Thus, \[ e^{2\theta}=3 \] Taking logarithm, \[ 2\theta=\log 3 \] \[ \theta=\frac{1}{2}\log 3 \]

Step 7: Final conclusion.
Therefore, \[ \boxed{\frac{1}{2}\log 3} \]