Current Attempt in Progress

With $x$ and $b$ in meters, a chain hangs in the shape of the catenary $coshx=\frac{1}{2}(e^x{e}^{-x})$ for $-bxb.$ If the chain is $12$ meters long, how far apart are its ends?

Round your answer to three decimal places.

The ends of the chain are

$$

meters apart.