To find the inverse function $f^{-1}(x)$ for the given function $f(x) = \left( 2e^x \right)^5$, let's follow these steps:

---

Step 1: Set $y = f(x)$

$y = \left( 2e^x \right)^5$

---

Step 2: Solve for $x$ in terms of $y$

Take the fifth root of both sides: $\sqrt[5]{y} = 2e^x$

Divide by 2: $\frac{\sqrt[5]{y}}{2} = e^x$

Take the natural logarithm ($\ln$) of both sides: $x = \ln\left( \frac{\sqrt[5]{y}}{2} \right)$

---

Step 3: Replace $x$ with $f^{-1}(x)$ and $y$ with $x$

$f^{-1}(x) = \ln\left( \frac{\sqrt[5]{x}}{2} \right)$

---

Final Answer:

$f^{-1}(x) = \ln\left( \frac{\sqrt[5]{x}}{2} \right)$

Let me know if you need further explanation or more details!

---

Questions to Explore:

1. What is the domain and range of $f(x)$ and its inverse $f^{-1}(x)$?
2. How would you verify that $f(f^{-1}(x)) = x$ and $f^{-1}(f(x)) = x$?
3. What is the behavior of $f(x)$ as $x \to \infty$ and $x \to -\infty$?
4. How would a change in the exponent (e.g., $2e^x$ raised to a different power) affect $f^{-1}(x)$?
5. Can you find the derivative of $f^{-1}(x)$ using the formula for the derivative of an inverse function?

---

Tip:

To verify an inverse function, compute both $f(f^{-1}(x))$ and $f^{-1}(f(x))$, ensuring both simplify to $x$.