Let's go through each part of the question, given that $g(x) = e^x$.

Part (a)

For $g(\ln 4)$: $g(\ln 4) = e^{\ln 4}$ Since $e^{\ln 4} = 4$, we have: $g(\ln 4) = 4$

Part (b)

For $g(\ln(5^2))$: $g(\ln(5^2)) = e^{\ln(5^2)}$ Using the property that $\ln(a^b) = b \ln a$, we get $\ln(5^2) = 2 \ln 5$. Therefore: $e^{\ln(5^2)} = e^{2 \ln 5} = (e^{\ln 5})^2 = 5^2 = 25$ Thus: $g(\ln(5^2)) = 25$

Part (c)

For $g\left(\ln\left(\frac{1}{e^4}\right)\right)$: $g\left(\ln\left(\frac{1}{e^4}\right)\right) = e^{\ln\left(\frac{1}{e^4}\right)}$ Using the property $\ln\left(\frac{1}{a}\right) = -\ln(a)$, we get: $\ln\left(\frac{1}{e^4}\right) = -\ln(e^4) = -4$ So: $e^{\ln\left(\frac{1}{e^4}\right)} = e^{-4}$ Thus: $g\left(\ln\left(\frac{1}{e^4}\right)\right) = e^{-4}$

Final Answers

1. $g(\ln 4) = 4$
2. $g(\ln(5^2)) = 25$
3. $g\left(\ln\left(\frac{1}{e^4}\right)\right) = e^{-4}$

Would you like further clarification on any part of this solution?

Here are some related questions to explore:

1. How does the property $e^{\ln x} = x$ apply in these scenarios?
2. What other properties of logarithms and exponentials might be useful here?
3. How would these calculations change if $g(x)$ had been defined differently?
4. Can you generalize these results for $g(\ln a)$ where $a$ is any positive number?
5. What other functions can be simplified similarly with $e$ and $\ln$ properties?

Tip: Remember that $e^{\ln x} = x$ is an essential property for simplifying expressions with $e$ and $\ln$.