Step 1: Write the given equation.
$2\cosh 2x + 10\sinh 2x = 5$.
Step 2: Express in terms of exponentials.
Recall $\cosh 2x = \frac{e^{2x}+e^{-2x}}{2}$ and $\sinh 2x = \frac{e^{2x}-e^{-2x}}{2}$.
Step 3: Substitute into the equation.
$2 \cdot \frac{e^{2x}+e^{-2x}}{2} + 10 \cdot \frac{e^{2x}-e^{-2x}}{2} = 5$
$(e^{2x}+e^{-2x}) + 5(e^{2x}-e^{-2x}) = 5$
$6e^{2x} - 4e^{-2x} = 5$.
Step 4: Let $t = e^{2x}$, so $e^{-2x} = 1/t$.
$6t - \frac{4}{t} = 5$
$6t^2 - 5t - 4 = 0$.
Step 5: Solve the quadratic.
Discriminant $= 25 + 96 = 121$.
$t = \frac{5 \pm 11}{12}$.
So $t = \frac{16}{12} = \frac{4}{3}$ or $t = \frac{-6}{12} = -\frac{1}{2}$. Since $t = e^{2x} > 0$, we take $t = \frac{4}{3}$.
Step 6: Solve for $x$.
$e^{2x} = \frac{4}{3}$, so $2x = \ln\frac{4}{3}$, giving $x = \frac{1}{2}\ln\frac{4}{3} = \frac{1}{2}\log\frac{4}{3}$. This is option (1).
\[ \boxed{x = \tfrac{1}{2}\log\tfrac{4}{3}} \]