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: Substitute \(x = 2u\) and expand using double angle hyperbolic formulas.
Using \(\cosh 2u = 1+2\sinh^2 u\) and \(\sinh 2u = 2\sinh u\cosh u\), the equation \(\cosh x - \frac{4}{5}\sinh x = 1\) becomes
\[ 1 + 2\sinh^2 u - \frac{4}{5}\cdot 2\sinh u\cosh u = 1 \]

Step 2: Simplify and factor.
The 1 cancels from both sides, leaving
\[ 2\sinh u\left(\sinh u - \frac{4}{5}\cosh u\right) = 0 \]

Step 3: Read off the two cases.
Either \(\sinh u = 0\), which gives \(u=0\), so \(x=0\) (the given solution), or
\[ \tanh u = \frac{4}{5} \]

Step 4: Solve for u using the inverse tanh formula.
\[ u = \tanh^{-1}\left(\frac{4}{5}\right) = \frac{1}{2}\log\left(\frac{1+\frac{4}{5}}{1-\frac{4}{5}}\right) = \frac{1}{2}\log 9 = \log 3 \]

Step 5: Recover x.
Since \(x=2u\), the second solution is
\[ \boxed{2\log 3} \]