The image you provided contains several math problems that involve finding the inverses of functions. Here's a breakdown of the problems in the image:

Problem 23:

$y = \left( \frac{e^{x+6}}{4} \right)^{\frac{1}{4}}$

Problem 24:

$y = \log_4 \left( \frac{e^{x+3}}{4} \right)$

To find the inverse of these functions, we need to swap the roles of $x$ and $y$ and then solve for the new $y$. Let me go step by step for each one:

For problem 23:

1. Start with $y = \left( \frac{e^{x+6}}{4} \right)^{\frac{1}{4}}$.
2. Raise both sides to the power of 4 to eliminate the fractional exponent: $y^4 = \frac{e^{x+6}}{4}$
3. Multiply both sides by 4: $4y^4 = e^{x+6}$
4. Take the natural logarithm (ln) of both sides: $\ln(4y^4) = x + 6$
5. Simplify the left-hand side: $\ln(4) + \ln(y^4) = x + 6$ $\ln(4) + 4\ln(y) = x + 6$
6. Solve for $x$: $x = \ln(4) + 4\ln(y) - 6$

The inverse function is: $x = \ln(4) + 4 \ln(y) - 6$

For problem 24:

1. Start with $y = \log_4 \left( \frac{e^{x+3}}{4} \right)$.
2. Convert the logarithmic equation to an exponential form: $4^y = \frac{e^{x+3}}{4}$
3. Multiply both sides by 4: $4^{y+1} = e^{x+3}$
4. Take the natural logarithm (ln) of both sides: $\ln(4^{y+1}) = x + 3$
5. Simplify using logarithmic properties: $(y+1)\ln(4) = x + 3$
6. Solve for $x$: $x = (y+1)\ln(4) - 3$

The inverse function is: $x = (y+1)\ln(4) - 3$

Would you like a more detailed explanation or have any specific questions about these steps? Here are some related questions for you to explore further:

1. How does taking the natural logarithm help solve for the inverse function?
2. What is the relationship between logarithms and exponents in these problems?
3. Can you explain the process of solving logarithmic equations more clearly?
4. How would the inverse functions change if different bases were used in the logarithms?
5. What other types of functions are commonly solved for inverses?

Tip: Practice with different bases and exponential forms to become more comfortable with solving for inverses in various scenarios!