The equation you're working with is:

$1.25 = \cosh\left(-\frac{2}{3}k\right)$

To solve for $k$, follow these steps:

Step 1: Isolate the hyperbolic cosine term

The hyperbolic cosine function, $\cosh(x)$, is even, meaning $\cosh(-x) = \cosh(x)$. Thus, the equation simplifies to:

$1.25 = \cosh\left(\frac{2}{3}k\right)$

Step 2: Use the inverse hyperbolic cosine

Apply the inverse hyperbolic cosine, $\cosh^{-1}(x)$, to both sides:

$\cosh^{-1}(1.25) = \frac{2}{3}k$

Step 3: Solve for $k$

Now, calculate $\cosh^{-1}(1.25)$ using a calculator or logarithmic identity:

$\cosh^{-1}(x) = \ln(x + \sqrt{x^2 - 1})$

Substituting $x = 1.25$:

$\cosh^{-1}(1.25) = \ln(1.25 + \sqrt{1.25^2 - 1})$

$= \ln(1.25 + \sqrt{1.5625 - 1}) = \ln(1.25 + \sqrt{0.5625}) = \ln(1.25 + 0.75) = \ln(2)$

$\cosh^{-1}(1.25) \approx \ln(2) \approx 0.6931$

Now, solve for $k$:

$0.6931 = \frac{2}{3}k$

Multiply both sides by $\frac{3}{2}$:

$k \approx \frac{3}{2} \times 0.6931 = 1.03965$

Thus, the value of $k$ is approximately:

$k \approx 1.04$

Would you like more details or have any questions?

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Here are some related questions:

1. How do you solve hyperbolic equations involving other functions like $\sinh(x)$?
2. What is the difference between hyperbolic and trigonometric functions?
3. Can hyperbolic functions be used to model real-world phenomena?
4. How can you express $\cosh(x)$ and $\sinh(x)$ in terms of exponentials?
5. What are the identities involving hyperbolic functions that parallel trigonometric ones?

Tip: When dealing with hyperbolic equations, always remember that $\cosh(x)$ is even, while $\sinh(x)$ is odd.