Question

If \[ 2\cosh2x+10\sinh2x=5, \] then \(x=\)

• A. \(\frac12\log\frac43\)
• B. \(\frac12\log\frac23\)
• C. \(\frac12\log\frac32\)
• D. \(\frac12\log\frac34\)

Hint:

For equations involving hyperbolic functions, convert \(\sinh x\) and \(\cosh x\) into exponential form using: \[ \cosh x=\frac{e^x+e^{-x}}{2} \] and \[ \sinh x=\frac{e^x-e^{-x}}{2}. \] Then solve the resulting algebraic equation.

Answer 1

The Correct Option is A

Step 1: Use hyperbolic function identities.
We know that \[ \cosh2x=\frac{e^{2x}+e^{-2x}}{2} \] and \[ \sinh2x=\frac{e^{2x}-e^{-2x}}{2} \] Substitute these into the given equation: \[ 2\left(\frac{e^{2x}+e^{-2x}}{2}\right) + 10\left(\frac{e^{2x}-e^{-2x}}{2}\right)=5 \]

Step 2: Simplify the equation.
\[ e^{2x}+e^{-2x}+5e^{2x}-5e^{-2x}=5 \] \[ 6e^{2x}-4e^{-2x}=5 \]

Step 3: Remove the negative exponent.
Multiply throughout by \[ e^{2x} \] \[ 6e^{4x}-4=5e^{2x} \]

Step 4: Substitute a variable.
Let \[ y=e^{2x} \] Then, \[ 6y^2-5y-4=0 \]

Step 5: Solve the quadratic equation.
Factorizing, \[ 6y^2-5y-4=0 \] \[ (3y-4)(2y+1)=0 \] So, \[ y=\frac43 \] or \[ y=-\frac12 \] But \[ y=e^{2x}\gt 0 \] Hence, \[ e^{2x}=\frac43 \]

Step 6: Find the value of \(x\).
Taking logarithm, \[ 2x=\log\frac43 \] Therefore, \[ x=\frac12\log\frac43 \] Hence, \[ \boxed{\frac12\log\frac43} \]