$$ \begin{array}{ll} ext { (7) } \frac{2^{x} \cdot \sqrt{2^{x}} \cdot \sqrt[5]{16^{x}}}{\sqrt[10]{8^{x}}} & ext { (8) } \frac{\sqrt{8^{x-2}} \cdot \sqrt[3]{25^{6 x+3}} \cdot \sqrt[3]{9^{x}}}{\sqrt[3]{6^{2 x-9}} \cdot 2^{-\frac{x}{6}}} \ ext { (b) Simplify the following expressions: } \ \begin{array}{ll} ext { (1) } \frac{3^{x+1}-4 \cdot 3^{x-1}}{3^{x-2}} & ext { (2) } \frac{8-2^{x+2}}{2^{x+1}-4}\end{array}\end{array} $$

Question

$$ \begin{array}{ll}	ext { (7) } \frac{2^{x} \cdot \sqrt{2^{x}} \cdot \sqrt[5]{16^{x}}}{\sqrt[10]{8^{x}}} & 	ext { (8) } \frac{\sqrt{8^{x-2}} \cdot \sqrt[3]{25^{6 x+3}} \cdot \sqrt[3]{9^{x}}}{\sqrt[3]{6^{2 x-9}} \cdot 2^{-\frac{x}{6}}} \ 	ext { (b) Simplify the following expressions: } \ \begin{array}{ll}	ext { (1) } \frac{3^{x+1}-4 \cdot 3^{x-1}}{3^{x-2}} & 	ext { (2) } \frac{8-2^{x+2}}{2^{x+1}-4}\end{array}\end{array}> $$

UpStudy AI · Accepted Answer

Let's simplify each expression step by step.

### Expression (7)
We have:
$$
\frac{2^{x} \cdot \sqrt{2^{x}} \cdot \sqrt[5]{16^{x}}}{\sqrt[10]{8^{x}}}
$$

1. **Convert all terms to base 2:**
   - $$ \sqrt{2^{x}} = 2^{x/2} $$
   - $$ 16^{x} = (2^4)^{x} = 2^{4x} $$ so $$ \sqrt[5]{16^{x}} = (2^{4x})^{1/5} = 2^{\frac{4x}{5}} $$
   - $$ 8^{x} = (2^3)^{x} = 2^{3x} $$ so $$ \sqrt[10]{8^{x}} = (2^{3x})^{1/10} = 2^{\frac{3x}{10}} $$

2. **Substituting back into the expression:**
   $$
   \frac{2^{x} \cdot 2^{x/2} \cdot 2^{\frac{4x}{5}}}{2^{\frac{3x}{10}}}
   $$

3. **Combine the exponents in the numerator:**
   $$
   2^{x + \frac{x}{2} + \frac{4x}{5}} = 2^{\frac{10x}{10} + \frac{5x}{10} + \frac{8x}{10}} = 2^{\frac{23x}{10}}
   $$

4. **Now simplify the entire expression:**
   $$
   \frac{2^{\frac{23x}{10}}}{2^{\frac{3x}{10}}} = 2^{\frac{23x}{10} - \frac{3x}{10}} = 2^{\frac{20x}{10}} = 2^{2x}
   $$

Thus, the simplified form of expression (7) is:
$$
2^{2x}
$$

### Expression (8)
We have:
$$
\frac{\sqrt{8^{x-2}} \cdot \sqrt[3]{25^{6x+3}} \cdot \sqrt[3]{9^{x}}}{\sqrt[3]{6^{2x-9}} \cdot 2^{-\frac{x}{6}}}
$$

1. **Convert all terms to base 2 and 3:**
   - $$ \sqrt{8^{x-2}} = \sqrt{(2^3)^{x-2}} = (2^{3(x-2)})^{1/2} = 2^{\frac{3(x-2)}{2}} = 2^{\frac{3x - 6}{2}} $$
   - $$ 25^{6x+3} = (5^2)^{6x+3} = 5^{12x + 6} $$ so $$ \sqrt[3]{25^{6x+3}} = (5^{12x + 6})^{1/3} = 5^{\frac{12x + 6}{3}} = 5^{4x + 2} $$
   - $$ 9^{x} = (3^2)^{x} = 3^{2x} $$ so $$ \sqrt[3]{9^{x}} = (3^{2x})^{1/3} = 3^{\frac{2x}{3}} $$
   - $$ 6^{2x-9} = (2 \cdot 3)^{2x-9} = 2^{2x-9} \cdot 3^{2x-9} $$ so $$ \sqrt[3]{6^{2x-9}} = (2^{2x-9} \cdot 3^{2x-9})^{1/3} = 2^{\frac{2x-9}{3}} \cdot 3^{\frac{2x-9}{3}} $$

2. **Substituting back into the expression:**
   $$
   \frac{2^{\frac{3x - 6}{2}} \cdot 5^{4x + 2} \cdot 3^{\frac{2x}{3}}}{2^{\frac{2x-9}{3}} \cdot 3^{\frac{2x-9}{3}} \cdot 2^{-\frac{x}{6}}}
   $$

3. **Combine the terms:**
   - For $$2$$:
   $$
   2^{\frac{3x - 6}{2} - \frac{2x-9}{3} + \frac{x}{6}}
   $$
   - For $$3$$:
   $$
   3^{\frac{2x}{3} - \frac{2x-9}{3}} = 3^{\frac{2x}{3} - \frac{2x}{3} + 3} = 3^{3}
   $$
   - For $$5$$:
   $$
   5^{4x + 2}
   $$

4. **Now simplify the exponent of $$2$$:**
   - Find a common denominator (6):
   $$
   \frac{3x - 6}{2} = \frac{9x - 18}{6}, \quad \frac{2x-9}{3} = \frac{4x - 18}{6}, \quad \frac{x}{6} = \frac{x}{6}
   $$
   - Combine:
   $$
   2^{\frac{9x - 18 - (4x - 18) + x}{6}} = 2^{\frac{9x - 18 - 4x + 18 + x}{6}} = 2^{\frac{6x}{6}} = 2^{x}
   $$

Thus, the simplified form of expression (8) is:
$$
2^{x} \cdot 3^{3} \cdot 5^{4x + 2}
$$

### Summary of Simplified Expressions
1. Expression (7): $$ 2^{2x} $$
2. Expression (8): $$ 2^{x} \cdot 3^{3} \cdot 5^{4x + 2} $$
Simplified Expressions: 1. $$ 2^{2x} $$ 2. $$ 2^{x} \cdot 3^{3} \cdot 5^{4x + 2} $$

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