find the value of k in this equation: 1.25=cosh(-2/3k)

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.

Learn how to solve the hyperbolic equation 1.25 = cosh(-2/3k) step by step. The solution involves key mathematical concepts such as hyperbolic functions and their inverse, with detailed explanations. Suitable for advanced high school students and college level.

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