Question

If \[ 4+6(e^{2x}+1)\tanh x = 11\cosh x+11\sinh x, \] then \[ x= \]

• A. \(\log 10\)
• B. \(\log 4\)
• C. \(\log 5\)
• D. \(\log 2\)

Hint:

Useful identities: \[ \cosh x+\sinh x=e^x \] and \[ \tanh x=\frac{e^{2x}-1}{e^{2x}+1} \] These simplify hyperbolic equations into algebraic equations.

Answer 1

The Correct Option is D

Step 1: Rewrite the hyperbolic functions in exponential form.
We know that \[ \tanh x=\frac{e^x-e^{-x}}{e^x+e^{-x}} \] Also, \[ \cosh x=\frac{e^x+e^{-x}}{2} \] and \[ \sinh x=\frac{e^x-e^{-x}}{2} \] Hence, \[ \cosh x+\sinh x = \frac{e^x+e^{-x}+e^x-e^{-x}}{2} \] \[ =e^x \] Therefore, the given equation becomes \[ 4+6(e^{2x}+1)\tanh x = 11e^x \]

Step 2: Simplify the \(\tanh x\) expression.
Using \[ \tanh x = \frac{e^{2x}-1}{e^{2x}+1}, \] we get \[ (e^{2x}+1)\tanh x = e^{2x}-1 \] Thus, \[ 4+6(e^{2x}-1)=11e^x \] \[ 4+6e^{2x}-6=11e^x \] \[ 6e^{2x}-2=11e^x \]

Step 3: Substitute \(e^x=t\).
Let \[ e^x=t \] Since \(e^x\gt 0\), \[ t\gt 0 \] Then, \[ 6t^2-11t-2=0 \]

Step 4: Solve the quadratic equation.
\[ 6t^2-11t-2=0 \] Factorizing, \[ 6t^2-12t+t-2=0 \] \[ 6t(t-2)+1(t-2)=0 \] \[ (6t+1)(t-2)=0 \] Thus, \[ t=2 \quad \text{or} \quad t=-\frac16 \] Since \[ t=e^x\gt 0, \] we reject \[ t=-\frac16 \] Hence, \[ e^x=2 \] Taking logarithm, \[ x=\log 2 \]

Step 5: Final conclusion.
Therefore, \[ \boxed{\log 2} \] which corresponds to option (4).