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The equation provided is:

\[
\frac{pr}{p} = \left(1 + \frac{\gamma - 1}{2} M^2 \right)^{\frac{\gamma}{\gamma - 1}}
\]

You are asked to solve for \(\gamma\). Let's go through the steps:

### Step 1: Simplify the Left Side
The left-hand side of the equation can be simplified as:
\[
\frac{pr}{p} = \frac{pr}{p} = \frac{p_r}{p}
\]

Thus, the equation becomes:
\[
\frac{p_r}{p} = \left(1 + \frac{\gamma - 1}{2} M^2 \right)^{\frac{\gamma}{\gamma - 1}}
\]

### Step 2: Isolate the Exponential Expression
Take the natural logarithm (ln) on both sides to eliminate the exponent on the right side. However, it would be more straightforward to express the solution using logarithms and algebra:

\[
\left(\frac{p_r}{p}\right)^{\frac{\gamma - 1}{\gamma}} = 1 + \frac{\gamma - 1}{2} M^2
\]

### Step 3: Rearrange the Equation
Now subtract 1 from both sides:

\[
\left(\frac{p_r}{p}\right)^{\frac{\gamma - 1}{\gamma}} - 1 = \frac{\gamma - 1}{2} M^2
\]

### Step 4: Solve for \(\gamma - 1\)
Multiply both sides by \(\frac{2}{M^2}\):

\[
\frac{2}{M^2} \left[ \left(\frac{p_r}{p}\right)^{\frac{\gamma - 1}{\gamma}} - 1 \right] = \gamma - 1
\]

### Step 5: Solve for \(\gamma\)
Finally, add 1 to both sides to get \(\gamma\):

\[
\gamma = \frac{2}{M^2} \left[ \left(\frac{p_r}{p}\right)^{\frac{\gamma - 1}{\gamma}} - 1 \right] + 1
\]

At this point, the equation is in a form where you can solve for \(\gamma\). This may require numerical methods or further simplifications depending on the specific values of \(M\), \(p_r\), and \(p\).

Would you like more details on solving it numerically or expanding on specific steps? Here are five related questions to consider:

1. What is the physical interpretation of the variables \(p_r\), \(p\), \(M\), and \(\gamma\) in the context of fluid dynamics?
2. How can numerical methods be applied to solve this equation for \(\gamma\)?
3. What is the significance of the ratio \(\frac{pr}{p}\) in thermodynamics?
4. Could this equation be simplified under certain assumptions about \(M\) or \(\gamma\)?
5. What are common values for \(\gamma\) in practical scenarios, such as air at different temperatures?

**Tip:** Always consider whether the equation can be simplified or if a numerical solution is necessary when dealing with complex exponents or logarithms.

Learn how to solve the equation involving isentropic flow relations to find the parameter gamma (γ). This problem involves fluid dynamics and thermodynamics concepts, specifically dealing with exponential equations and is suitable for advanced learners and professionals.

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