a.) $$ \frac{6^{x} \cdot 9^{x+1} \cdot 2}{27^{x+1} \cdot 2^{x-1}} $$

Question

UpStudy AI · Accepted Answer

To simplify the expression:

$$
\frac{6^{x} \cdot 9^{x+1} \cdot 2}{27^{x+1} \cdot 2^{x-1}}
$$

Let's break it down step by step by expressing each number in terms of its prime factors:

1. **Expressing in Prime Factors:**
   - $$6 = 2 \times 3$$
   - $$9 = 3^2$$
   - $$27 = 3^3$$

2. **Rewrite the Expression:**
   
   Substitute the prime factors into the original expression:

$$
   \frac{(2 \cdot 3)^{x} \cdot (3^2)^{x+1} \cdot 2}{(3^3)^{x+1} \cdot 2^{x-1}}
   $$

3. **Apply Exponent Rules:**

- $$(a \cdot b)^n = a^n \cdot b^n$$
   - $$(a^m)^n = a^{m \cdot n}$$

Applying these rules:

$$
   \frac{2^{x} \cdot 3^{x} \cdot 3^{2(x+1)} \cdot 2}{3^{3(x+1)} \cdot 2^{x-1}}
   $$

Simplify the exponents:

$$
   \frac{2^{x} \cdot 2 \cdot 3^{x} \cdot 3^{2x + 2}}{3^{3x + 3} \cdot 2^{x - 1}}
   $$

4. **Combine Like Terms:**

- For the base 2 terms:

$$
     2^{x} \cdot 2 = 2^{x+1}
     $$

$$
     \frac{2^{x+1}}{2^{x-1}} = 2^{(x+1) - (x-1)} = 2^2 = 4
     $$

- For the base 3 terms:

$$
     3^{x} \cdot 3^{2x + 2} = 3^{3x + 2}
     $$

$$
     \frac{3^{3x + 2}}{3^{3x + 3}} = 3^{(3x + 2) - (3x + 3)} = 3^{-1} = \frac{1}{3}
     $$

5. **Final Simplification:**

Combine the simplified base 2 and base 3 terms:

$$
   4 \times \frac{1}{3} = \frac{4}{3}
   $$

**Final Answer:**

$$
\frac{4}{3}
$$
$$ \frac{4}{3} $$

a.) \( \frac{6^{x} \cdot 9^{x+1} \cdot 2}{27^{x+1} \cdot 2^{x-1}} \)