$$ \begin{array}{ll} ext { d) } \frac{81^{a+1} \cdot 9^{3-a}}{7^{a+2} \cdot 3^{1-a}} & ext { e) } \frac{32^{x} \cdot 2^{2-x}}{8^{x+1} \cdot 2^{x}}\end{array} $$

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$$ \begin{array}{ll}	ext { d) } \frac{81^{a+1} \cdot 9^{3-a}}{7^{a+2} \cdot 3^{1-a}} & 	ext { e) } \frac{32^{x} \cdot 2^{2-x}}{8^{x+1} \cdot 2^{x}}\end{array} $$

UpStudy AI · Accepted Answer

Let's simplify each expression step by step.

---

### **d)**
$$
\frac{81^{a+1} \cdot 9^{3-a}}{7^{a+2} \cdot 3^{1-a}}
$$

**Step 1: Express all terms with base 3.**
- $$81 = 3^4$$
- $$9 = 3^2$$

So,
$$
81^{a+1} = (3^4)^{a+1} = 3^{4(a+1)} = 3^{4a + 4}
$$
$$
9^{3-a} = (3^2)^{3-a} = 3^{2(3-a)} = 3^{6 - 2a}
$$

**Step 2: Combine the numerator.**
$$
3^{4a + 4} \cdot 3^{6 - 2a} = 3^{(4a + 4) + (6 - 2a)} = 3^{2a + 10}
$$

**Step 3: Combine the denominator.**
$$
7^{a+2} \cdot 3^{1 - a}
$$

**Step 4: Simplify the entire expression.**
$$
\frac{3^{2a + 10}}{7^{a+2} \cdot 3^{1 - a}} = \frac{3^{2a + 10}}{3^{1 - a}} \cdot \frac{1}{7^{a+2}} = 3^{(2a + 10) - (1 - a)} \cdot 7^{-(a+2)} = 3^{3a + 9} \cdot 7^{-(a+2)}
$$
$$
= \frac{3^{3a + 9}}{7^{a + 2}}
$$

---

### **e)**
$$
\frac{32^{x} \cdot 2^{2-x}}{8^{x+1} \cdot 2^{x}}
$$

**Step 1: Express all terms with base 2.**
- $$32 = 2^5$$
- $$8 = 2^3$$

So,
$$
32^{x} = (2^5)^x = 2^{5x}
$$
$$
8^{x+1} = (2^3)^{x+1} = 2^{3(x+1)} = 2^{3x + 3}
$$

**Step 2: Combine the numerator.**
$$
2^{5x} \cdot 2^{2 - x} = 2^{5x + 2 - x} = 2^{4x + 2}
$$

**Step 3: Combine the denominator.**
$$
2^{3x + 3} \cdot 2^{x} = 2^{3x + 3 + x} = 2^{4x + 3}
$$

**Step 4: Simplify the entire expression.**
$$
\frac{2^{4x + 2}}{2^{4x + 3}} = 2^{(4x + 2) - (4x + 3)} = 2^{-1} = \frac{1}{2}
$$

---

### **Final Simplified Forms:**

- **d)** $$\displaystyle \frac{3^{3a + 9}}{7^{a + 2}}$$
- **e)** $$\displaystyle \frac{1}{2}$$
The simplified forms are: - **d)** $$\frac{3^{3a + 9}}{7^{a + 2}}$$ - **e)** $$\frac{1}{2}$$

\( \begin{array}{ll}\text { d) } \frac{81^{a+1} \cdot 9^{3-a}}{7^{a+2} \cdot 3^{1-a}} & \text { e) } \frac{32^{x} \cdot 2^{2-x}}{8^{x+1} \cdot 2^{x}}\end{array} \)

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