Let's analyze and simplify the given expressions.

Problem 37

Expression:
$\frac{121^{3n}}{22^{2n}}$

Step 1: Rewrite the bases in terms of prime factors.

• $121 = 11^2$, so $121^{3n} = (11^2)^{3n} = 11^{6n}$.
• $22 = 2 \cdot 11$, so $22^{2n} = (2 \cdot 11)^{2n} = 2^{2n} \cdot 11^{2n}$.

Now the expression becomes:
$\frac{11^{6n}}{2^{2n} \cdot 11^{2n}}$

Step 2: Simplify the powers.
Using the property $a^m / a^n = a^{m-n}$:
$\frac{11^{6n}}{11^{2n}} = 11^{(6n - 2n)} = 11^{4n}.$

The result is:
$\frac{11^{4n}}{2^{2n}}$

Final Answer for Problem 37:
$\frac{11^{4n}}{2^{2n}}$

Problem 38

Expression:
$\frac{4^{2n} \cdot 8^{3n}}{16^{4n}}$

Step 1: Rewrite the bases in terms of powers of 2.

• $4 = 2^2$, so $4^{2n} = (2^2)^{2n} = 2^{4n}$.
• $8 = 2^3$, so $8^{3n} = (2^3)^{3n} = 2^{9n}$.
• $16 = 2^4$, so $16^{4n} = (2^4)^{4n} = 2^{16n}$.

Now the expression becomes:
$\frac{2^{4n} \cdot 2^{9n}}{2^{16n}}$

Step 2: Combine powers in the numerator.
Using $a^m \cdot a^n = a^{m+n}$:
$2^{4n} \cdot 2^{9n} = 2^{(4n + 9n)} = 2^{13n}.$

The expression is now:
$\frac{2^{13n}}{2^{16n}}$

Step 3: Simplify the fraction.
Using $a^m / a^n = a^{m-n}$:
$\frac{2^{13n}}{2^{16n}} = 2^{(13n - 16n)} = 2^{-3n}.$

Final Answer for Problem 38:
$2^{-3n} = \frac{1}{2^{3n}}$

Summary of Answers:

1. Problem 37: $\frac{11^{4n}}{2^{2n}}$
2. Problem 38: $\frac{1}{2^{3n}}$

Would you like me to expand the steps further or solve other problems?

Follow-Up Questions:

1. How are the properties of exponents applied to simplify fractional powers?
2. What happens when bases are expressed in terms of primes during simplification?
3. Why does simplifying the numerator and denominator separately often help?
4. How can negative exponents be interpreted in fraction form?
5. Why is factoring important in problems like these?

Tip: When working with exponents, always break down composite bases into their prime factors to simplify calculations!