1. Simplify each expression: i. $(x^5)^1 = x^5$ because any power raised to 1 remains the same. ii. $(x^1)^2 \times x^1 = x^{1 \times 2} \times x^1 = x^2 \times x^1 = x^{2+1} = x^3$ iii. $x^2 + x^8$ cannot be simplified further because terms have different exponents. iv. $\sqrt{x^{10}} = x^{10/2} = x^5$ v. $(x^1)^5 = x^{1 \times 5} = x^5$ vi. $\sqrt[3]{x^5} = x^{5/3}$ vii. $(x^{-1})^2 \times (x^1)^3 = x^{-2} \times x^3 = x^{-2+3} = x^1 = x$ viii. $3x^2 \times 5x^4 y^3 = (3 \times 5) x^{2+4} y^3 = 15x^6 y^3$ ix. $\sqrt{25} \times x^5 \times y^{-4} = 5 x^5 y^{-4}$ x. $\sqrt{9} \times x^1 \times y^{-1} \times \sqrt{8} \times x^6 \times y^{-3} = 3x y^{-1} \times 2\sqrt{2} x^6 y^{-3} = 6\sqrt{2} x^{1+6} y^{-1-3} = 6\sqrt{2} x^7 y^{-4}$ 2. Simplify the roots expressions: i. $\sqrt{20} + \sqrt{5} = \sqrt{4 \times 5} + \sqrt{5} = 2\sqrt{5} + \sqrt{5} = 3\sqrt{5}$ ii. $\sqrt{24x} - \sqrt{81x} = \sqrt{4 \times 6 x} - \sqrt{9^2 x} = 2\sqrt{6x} - 9\sqrt{x}$. Since $\sqrt{6x} \neq \sqrt{x}$, this is simplified form. iii. $\sqrt{4} y^{-1} \times \sqrt{12} y^{1} = 2 y^{-1} \times 2\sqrt{3} y^{1} = 4 \sqrt{3} y^{0} = 4\sqrt{3}$ iv. $\sqrt[3]{\frac{3}{32}} = \frac{\sqrt[3]{3}}{\sqrt[3]{32}} = \frac{\sqrt[3]{3}}{2^{5/3}}$ v. $\sqrt[3]{\frac{8}{1000}} = \frac{\sqrt[3]{8}}{\sqrt[3]{1000}} = \frac{2}{10} = \frac{1}{5}$ vi. $\sqrt[4]{16^1} = (16)^{1/4} = 2$ 3. Find integers $x$ and $y$ such that: $$2^x \times 3^y = 6^4$$ Rewrite RHS: $$6^4 = (2 \times 3)^4 = 2^4 \times 3^4$$ Since bases match, equate exponents: $$x = 4, \quad y = 4$$ 4. Find $x$: i. $$\frac{3^3 \times 3^5}{3^7 \times 3^3} = 3^x$$ Simplify numerator: $3^{3+5} = 3^8$ Simplify denominator: $3^{7+3} = 3^{10}$ Overall: $3^{8-10} = 3^{-2} = 3^x$ so $x = -2$ ii. $$\frac{(7^1 \times 7^3)^2}{7^1 + 7^2} = 7^1$$ Simplify numerator: $(7^{1+3})^2 = (7^4)^2 = 7^{8}$ Denominator: $7^1 + 7^2$ does not simplify to a single power of 7, so reinterpret statement or typographic error; assuming denominator meant multiplication: If denominator is $7^1 \times 7^2 = 7^{3}$ then: $$\frac{7^{8}}{7^{3}} = 7^{8-3} = 7^{5} \neq 7^{1}$$ Thus, $x = 5$ is not consistent with target $7^1$, so likely denominator is sum. Since denominator is sum, LHS is not a pure power of 7; equation has no solution for $x$ as scalar exponent. iii. $$4^x = \frac{1}{64}$$ Write $4=(2^2)$ and $64 = 2^6$: $$ (2^2)^x = 2^{-6} \quad \Rightarrow \quad 2^{2x} = 2^{-6}$$ Equate exponents: $$2x = -6 \Rightarrow x = -3$$ iv. $$2^x = 0.125$$ Recognize $0.125 = \frac{1}{8} = 2^{-3}$ So $x = -3$ 5. Given $$\frac{(36x^1)^2}{8x^2 \times 3x} = 2^3 x^1$$ Simplify numerator: $$36^2 x^{2} = 1296 x^2$$ Simplify denominator: $$8x^2 \times 3 x = 24 x^{3}$$ Fraction: $$\frac{1296 x^2}{24 x^{3}} = \frac{1296}{24} x^{2-3} = 54 x^{-1}$$ According to problem: $$54 x^{-1} = 2^3 x^1 = 8 x^1$$ Equate coefficients: $$54 = 8$$ Not equal, contradiction unless assuming constants separated from $x$ terms: Equate powers of $x$: $$x^{-1} = x^{1} \Rightarrow -1 = 1$$ impossible. So consider the problem may require rewriting or finding $a,b,c$ in expression $a^b x^c$ matching both sides. Let us write $$54 x^{-1} = a^b x^c \Rightarrow a^b = 54, c = -1$$ Given RHS is $2^3 x^1 = 8 x^1$ not equal, so solution is $a=54^{1/b}, b \text{ free}, c = -1$. Problem likely incomplete or need clarification. 6. Given $$\frac{\sqrt{x^{-1}} \times \sqrt{y^2}}{\sqrt{x^{6} y^{2/3}}} = x^1 y^1$$ Simplify numerator: $$x^{-1/2} y^{1}$$ Simplify denominator: $$\sqrt{x^{6} y^{2/3}} = x^{6/2} y^{(2/3)/2} = x^3 y^{1/3}$$ Overall: $$\frac{x^{-1/2} y^{1}}{x^3 y^{1/3}} = x^{-1/2 - 3} y^{1 - 1/3} = x^{-7/2} y^{2/3}$$ According to problem: $$x^1 y^1 = x^{-7/2} y^{2/3}$$ Equate exponents: $$1 = -\frac{7}{2} \quad \Rightarrow \text{contradiction}$$ Similarly $1 = \frac{2}{3}$ no. So likely problem aimed to find $a,b$ such that original expression equals $x^a y^b$. Then $$a = -\frac{7}{2}, b = \frac{2}{3}$$ 7. Given $$\frac{\sqrt{a^{3/2} b^{3/2}}}{a^{1/2} b^{3/2}} = a^1 b^1$$ Simplify numerator: $$\sqrt{a^{3/2} b^{3/2}} = a^{3/4} b^{3/4}$$ Simplify fraction: $$\frac{a^{3/4} b^{3/4}}{a^{1/2} b^{3/2}} = a^{3/4 - 1/2} b^{3/4 - 3/2} = a^{1/4} b^{-3/4}$$ Given equals $a^1 b^1$: Equate exponents: $$\frac{1}{4} = 1 \Rightarrow \text{false}$$ and $$-\frac{3}{4} = 1 \Rightarrow \text{false}$$ So problem asks to find $x,y$ such that the expression equals $a^x b^y$, conclude $$x = \frac{1}{4}, y = -\frac{3}{4}$$ Final Answers: 1. i.$x^5$ ii.$x^3$ iii.$x^2 + x^8$ iv.$x^5$ v.$x^5$ vi.$x^{5/3}$ vii.$x$ viii.$15x^6 y^3$ ix.$5 x^5 y^{-4}$ x.$6 \sqrt{2} x^7 y^{-4}$ 2. i.$3 \sqrt{5}$ ii.$2 \sqrt{6x} - 9 \sqrt{x}$ iii.$4 \sqrt{3}$ iv.$\frac{\sqrt[3]{3}}{2^{5/3}}$ v.$\frac{1}{5}$ vi.$2$ 3. $x=4, y=4$ 4. i.$x=-2$ ii. no solution (as sum in denominator) iii.$x=-3$ iv.$x=-3$ 5. $a=54^{1/b}, c=-1$, problem incomplete 6. $a=-\frac{7}{2}, b=\frac{2}{3}$ 7. $x=\frac{1}{4}, y=-\frac{3}{4}$