Question

Solve for $ x $ given: $$ 2 \cosh(2x) + 10 \sinh(2x) = 5 $$

• A. \( \frac{1}{2} \log\left( \frac{4}{3} \right) \)
• B. \( \frac{1}{2} \log\left( \frac{2}{3} \right) \)
• C. \( \frac{1}{2} \log\left( \frac{3}{2} \right) \)
• D. \( \frac{1}{2} \log\left( \frac{3}{4} \right) \)

Hint:

Use exponential definitions of hyperbolic functions to convert the equation into a solvable quadratic in \( e^{2x} \).

Answer 1

The Correct Option is A

Recall the identities: \[ \cosh(2x) = \frac{e^{2x} + e^{-2x}}{2}, \quad \sinh(2x) = \frac{e^{2x} - e^{-2x}}{2} \] Substitute into the equation: \[ 2 \cdot \frac{e^{2x} + e^{-2x}}{2} + 10 \cdot \frac{e^{2x} - e^{-2x}}{2} = 5 \Rightarrow (e^{2x} + e^{-2x}) + 5(e^{2x} - e^{-2x}) = 5 \] Simplify: \[ e^{2x} + e^{-2x} + 5e^{2x} - 5e^{-2x} = 5 \Rightarrow (1 + 5)e^{2x} + (1 - 5)e^{-2x} = 5 \Rightarrow 6e^{2x} - 4e^{-2x} = 5 \] Multiply through by \( e^{2x} \): \[ 6e^{4x} - 4 = 5e^{2x} \Rightarrow 6e^{4x} - 5e^{2x} - 4 = 0 \] Let \( y = e^{2x} \), then: \[ 6y^2 - 5y - 4 = 0 \Rightarrow y = \frac{5 \pm \sqrt{25 + 96}}{12} = \frac{5 \pm \sqrt{121}}{12} = \frac{5 \pm 11}{12} \] So: \[ y = \frac{16}{12} = \frac{4}{3},\quad \text{or } y = \frac{-6}{12} = -\frac{1}{2} \text{ (reject, since } y = e^{2x}>0\text{)} \] Now: \[ e^{2x} = \frac{4}{3} \Rightarrow 2x = \log\left(\frac{4}{3}\right) \Rightarrow x = \frac{1}{2} \log\left( \frac{4}{3} \right) \]