Question

If one solution of the equation \[ \cosh x - \frac{4}{5} \sinh x = 1 \text{ is } x = 0, \] then the other solution is:

• A. \(2 \log 2\)
• B. \(2 \log 5\)
• C. \(\log \frac{4}{3}\)
• D. \(2 \log 3\)

Hint:

Convert hyperbolic functions to exponentials to solve linear combinations; solve the resulting quadratic in \(e^x\).

Answer 1

The Correct Option is D

Step 1: Start with the given equation.
\[ \cosh x - \frac{4}{5} \sinh x = 1 \]

Step 2: Express in terms of exponentials.
\[ \cosh x = \frac{e^x + e^{-x}}{2}, \quad \sinh x = \frac{e^x - e^{-x}}{2} \] \[ \frac{e^x + e^{-x}}{2} - \frac{4}{5} \cdot \frac{e^x - e^{-x}}{2} = 1 \]

Step 3: Simplify equation.
\[ \frac{e^x + e^{-x} - \frac{4}{5} (e^x - e^{-x})}{2} = 1 \] \[ \frac{(1-\frac{4}{5}) e^x + (1+\frac{4}{5}) e^{-x}}{2} = 1 \] \[ \frac{\frac{1}{5} e^x + \frac{9}{5} e^{-x}}{2} = 1 \] \[ e^x + 9 e^{-x} = 10 \]

Step 4: Multiply both sides by \(e^x\).
\[ e^{2x} + 9 = 10 e^x \Rightarrow e^{2x} - 10 e^x + 9 = 0 \]

Step 5: Solve quadratic in \(e^x\).
\[ e^x = \frac{10 \pm \sqrt{100 - 36}}{2} = \frac{10 \pm 8}{2} = 9 \text{ or } 1 \]

Step 6: Find \(x\).
- \(x = \log 1 = 0\) (given)
- \(x = \log 9 = 2 \log 3\)
\[ \boxed{x = 2 \log 3} \]