1. Simplify the expression $ \frac{x^7}{x^2 y^3} $. Using the Quotient Rule for exponents: $ \frac{b^m}{b^n} = b^{m-n} $, we get: $$\frac{x^7}{x^2 y^3} = x^{7-2} y^{-3} = x^5 y^{-3} = \frac{x^5}{y^3}$$ 2. Simplify $ 2.5^{2-3y} $. This is already in simplest exponential form. It can be written as: $$2.5^{2-3y}$$ 3. Evaluate the true statements: - $ \frac{2^{1/2}}{2^{3y}} = 2^{1/2 - 3y} $, not $ 2^{3(y-x)} $, so false. - $ 8^{3x - 1/3} = (2^3)^{3x - 1/3} = 2^{9x - 1} $. - $ \frac{512^x}{2} = \frac{(2^9)^x}{2} = \frac{2^{9x}}{2} = 2^{9x - 1} $. So, $ 8^{3x - 1/3} = \frac{512^x}{2} $ is true. - $ 6^x \neq \frac{6x}{6y} $, false. - $ 0.5^{2x + 3} = (\frac{1}{2})^{2x + 3} = 2^{-(2x + 3)} = 2^{-2x - 3} $. - $ 0.25^{x + 3} = (\frac{1}{4})^{x + 3} = 4^{-(x + 3)} = (2^2)^{-(x + 3)} = 2^{-2x - 6} $. Since $ 2^{-2x - 3} \neq 2^{-2x - 6} $, false. - $ 5^{2x - y} = 5^{2x} \cdot 5^{-y} = 25^x \cdot 5^{-y} $, not $ 25^x \cdot 5^y $, false. 4. Express $ 8^x \cdot 6^2 \cdot (\frac{3}{2})^{-4} $ as $ 2^a \cdot 3^b $. Rewrite bases: - $ 8^x = (2^3)^x = 2^{3x} $ - $ 6^2 = (2 \cdot 3)^2 = 2^2 \cdot 3^2 $ - $ (\frac{3}{2})^{-4} = (\frac{2}{3})^4 = 2^4 \cdot 3^{-4} $ Multiply all: $$2^{3x} \cdot 2^2 \cdot 3^2 \cdot 2^4 \cdot 3^{-4} = 2^{3x + 2 + 4} \cdot 3^{2 - 4} = 2^{3x + 6} \cdot 3^{-2}$$ So, $ a = 3x + 6 $ and $ b = -2 $. --- **Final answers:** - $ \frac{x^7}{x^2 y^3} = \frac{x^5}{y^3} $ - $ 2.5^{2-3y} $ (already simplified) - True statement: $ 8^{3x - 1/3} = \frac{512^x}{2} $ - $ a = 3x + 6 $, $ b = -2 $