1. Simplify $2^4 \cdot 2^{-1}$. Using the exponent rule $a^m \cdot a^n = a^{m+n}$: $$2^4 \cdot 2^{-1} = 2^{4 + (-1)} = 2^3 = 8$$ 2. Simplify $3^0 \cdot 3^3$. Recall $a^0 = 1$ for $a \neq 0$: $$3^0 \cdot 3^3 = 1 \cdot 27 = 27$$ 3. Simplify $(2a^2 - b^{-2})^{-3}$. Rewrite $b^{-2} = \frac{1}{b^2}$, expression inside parentheses is $2a^2 - \frac{1}{b^2}$, but since the problem only asks to simplify the expression with exponents, and no further factorization is provided, final form is: $$(2a^2 - b^{-2})^{-3}$$ (simplified as given). 4. Simplify $(18a^{2}b^{-2} - 7a^{3}b^{-5})^{0}$. Any non-zero expression raised to zero is 1: $$= 1$$ 5. Simplify $3c^{-1}d^{-1}(2cd - 7c^2 d^2)$. Rewrite negative exponents: $c^{-1} = \frac{1}{c}$, $d^{-1} = \frac{1}{d}$ So: $$3 \cdot \frac{1}{c} \cdot \frac{1}{d} (2cd - 7c^2 d^2) = 3 \cdot \frac{1}{cd} (2cd - 7c^2 d^2)$$ Distribute: $$= 3 \left(\frac{2cd}{cd} - \frac{7c^2 d^2}{cd}\right) = 3 (2 - 7c d) = 6 - 21 c d$$ 6. Simplify $(a - 2b^3 c)^{-3}$. This is already simplified; expressing negative exponent as reciprocal: $$= \frac{1}{(a - 2b^3 c)^3}$$ 7. Simplify $a b^{-2} - a^{-2} b$. Rewrite as: $$\frac{a}{b^2} - \frac{b}{a^2}$$ Common denominator is $a^2 b^2$, so: $$= \frac{a \cdot a^2}{a^2 b^2} - \frac{b \cdot b^2}{a^2 b^2} = \frac{a^3 - b^3}{a^2 b^2}$$ 8. Simplify $a x^{-2} - 5 y^{-9}$. Rewrite negative exponents: $$\frac{a}{x^2} - \frac{5}{y^9}$$ 9. Simplify $\frac{2m}{n^{-2}} + \frac{n^3}{2^{-1} m^{-1}}$. Rewrite exponents: $$\frac{2m}{n^{-2}} = 2m \cdot n^2$$ $$\frac{n^3}{2^{-1} m^{-1}} = n^3 \cdot 2^1 \cdot m^{1} = 2 m n^3$$ Add: $$2 m n^2 + 2 m n^3 = 2 m n^2 (1 + n)$$ 10. Simplify $\frac{5}{a^{-2} + b^{-2}}$. Rewrite: $$\frac{5}{\frac{1}{a^2} + \frac{1}{b^2}} = \frac{5}{\frac{b^2 + a^2}{a^2 b^2}} = 5 \cdot \frac{a^2 b^2}{a^2 + b^2}$$ 11. Simplify $\frac{7}{x^{-1}}$. Rewrite denominator: $$x^{-1} = \frac{1}{x}$$ Thus: $$\frac{7}{\frac{1}{x}} = 7x$$ 12. Simplify $\frac{(ab)^{-1}}{(a^{-1} b^{-1})}$. Rewrite: $$\frac{1/(ab)}{1/a \cdot 1/b} = \frac{1/(ab)}{1/(ab)} = 1$$ 13. Simplify $\frac{(m^2 n^3)^{-3}}{(m n)^{-2}}$. Rewrite: $$(m^2 n^3)^{-3} = m^{-6} n^{-9}$$ $$(m n)^{-2} = m^{-2} n^{-2}$$ Divide: $$\frac{m^{-6} n^{-9}}{m^{-2} n^{-2}} = m^{-6 - (-2)} n^{-9 - (-2)} = m^{-4} n^{-7} = \frac{1}{m^4 n^7}$$ 14. Simplify $\frac{(2a)^{-2} b^2}{(8a)^{-1} b c^0}$. Recall $c^0 = 1$. Rewrite: $(2a)^{-2} = 2^{-2} a^{-2} = \frac{1}{4 a^2}$ $(8a)^{-1} = \frac{1}{8a}$ So: $$\frac{\frac{1}{4 a^2} b^2}{\frac{1}{8 a} b} = \frac{b^2}{4 a^2} \cdot \frac{8 a}{b} = \frac{b^2}{4 a^2} \cdot \frac{8 a}{b} = \frac{8 a b^2}{4 a^2 b} = \frac{8 b}{4 a} = 2 \frac{b}{a}$$ 15. Simplify $\frac{4 a^3 b^2 c^0}{2 a b^{-1}}$. Rewrite: $c^0 = 1$, $b^{-1} = \frac{1}{b}$ So denominator: $2 a \cdot \frac{1}{b} = \frac{2 a}{b}$ Fraction: $$\frac{4 a^3 b^2}{\frac{2 a}{b}} = 4 a^3 b^2 \cdot \frac{b}{2 a} = \frac{4}{2} a^{3-1} b^{2+1} = 2 a^{2} b^{3}$$ 16. Simplify $(3 r s - 2 p^3)^{-4}$. This is fully simplified; written as reciprocal: $$= \frac{1}{(3 r s - 2 p^3)^4}$$ 17. Simplify $(8 s^{-1} t^{4})(2^{-1} s + 2)$. Rewrite: $8 s^{-1} t^4 = 8 \cdot \frac{1}{s} t^4 = \frac{8 t^4}{s}$ $2^{-1} = \frac{1}{2}$ Inside parentheses: $$\frac{1}{2} s + 2 = \frac{s}{2} + 2$$ Multiply: $$\frac{8 t^4}{s} \left( \frac{s}{2} + 2 \right) = \frac{8 t^4}{s} \cdot \frac{s}{2} + \frac{8 t^4}{s} \cdot 2 = 4 t^4 + \frac{16 t^4}{s}$$ 18. Simplify $a b^{-1} + a^{-1} b$. Rewrite: $$\frac{a}{b} + \frac{b}{a} = \frac{a^2 + b^2}{ab}$$ 19. Simplify $(3 x^{2} + y^{3})^{-2}$. Already simplified; write reciprocal squared: $$= \frac{1}{(3 x^{2} + y^{3})^{2}}$$ 20. Simplify $48 a b^{-1} c (4 a b c^0)^{-2}$. Rewrite $b^{-1} = \frac{1}{b}, c^{0} = 1$ Inside parentheses: $$4 a b \cdot 1 = 4 a b$$ Raised to $-2$: $$(4 a b)^{-2} = 4^{-2} a^{-2} b^{-2} = \frac{1}{16 a^{2} b^{2}}$$ Multiply everything: $$48 a \cdot \frac{1}{b} \cdot c \cdot \frac{1}{16 a^{2} b^{2}} = \frac{48 a c}{b} \cdot \frac{1}{16 a^{2} b^{2}} = \frac{48 c}{16 a b^{3}} = \frac{3 c}{a b^{3}}$$ 21. Simplify $(x + y)^{-1} (x + y)$. This is: $$\frac{1}{x + y} (x + y) = 1$$ 22. Simplify $x y (x^{-1} + y^{-7})$. Rewrite: $$x y \left( \frac{1}{x} + \frac{1}{y^{7}} \right) = y + \frac{x y}{y^{7}} = y + x y^{1 - 7} = y + x y^{-6} = y + \frac{x}{y^{6}}$$ 23. Simplify $(6 a^{3})^{2} (12 a^{4})^{-1}$. Calculate each term: $$(6 a^{3})^{2} = 6^{2} a^{6} = 36 a^{6}$$ $$(12 a^{4})^{-1} = \frac{1}{12 a^{4}}$$ Multiply: $$36 a^{6} \cdot \frac{1}{12 a^{4}} = 3 a^{2}$$ 24. Simplify $\frac{4 a^{2} b^{-3}}{8 a^{-5} b^{2}}$. Rewrite $b^{-3} = \frac{1}{b^{3}}, a^{-5} = \frac{1}{a^{5}}$: $$\frac{4 a^{2} / b^{3}}{8 / a^{5} b^{2}} = \frac{4 a^{2}}{b^{3}} \cdot \frac{a^{5} b^{2}}{8} = \frac{4}{8} a^{2 + 5} b^{2 - 3} = \frac{1}{2} a^{7} b^{-1} = \frac{a^{7}}{2 b}$$ 25. Simplify $\frac{4^{3} \cdot 16^{4} \cdot 2^{5}}{8^{6} \cdot 32^{2}}$. Rewrite bases as powers of 2: $$4 = 2^{2}, 16 = 2^{4}, 8 = 2^{3}, 32 = 2^{5}$$ Calculate exponents: $$4^{3} = (2^{2})^{3} = 2^{6}$$ $$16^{4} = (2^{4})^{4} = 2^{16}$$ $$8^{6} = (2^{3})^{6} = 2^{18}$$ $$32^{2} = (2^{5})^{2} = 2^{10}$$ Now the fraction becomes: $$\frac{2^{6} \cdot 2^{16} \cdot 2^{5}}{2^{18} \cdot 2^{10}} = \frac{2^{27}}{2^{28}} = 2^{27 - 28} = 2^{-1} = \frac{1}{2}$$ 26. Simplify $\frac{2 x^{-4}}{3 x^{-2}}$. Rewrite: $$\frac{2 / x^{4}}{3 / x^{2}} = \frac{2}{x^{4}} \cdot \frac{x^{2}}{3} = \frac{2 x^{2}}{3 x^{4}} = \frac{2}{3 x^{2}}$$ 27. Simplify $\frac{4^{-1} m^{-2} n^{-3}}{(2 m n)^{-3}}$. Rewrite: $$4^{-1} = \frac{1}{4}, m^{-2} = \frac{1}{m^{2}}, n^{-3} = \frac{1}{n^{3}}$$ $$(2 m n)^{-3} = 2^{-3} m^{-3} n^{-3} = \frac{1}{8 m^{3} n^{3}}$$ So, $$\frac{\frac{1}{4 m^{2} n^{3}}}{\frac{1}{8 m^{3} n^{3}}} = \frac{1}{4 m^{2} n^{3}} \cdot \frac{8 m^{3} n^{3}}{1} = \frac{8 m^{3} n^{3}}{4 m^{2} n^{3}} = 2 m$$ 28. Simplify $\frac{2 a - b^{0}}{(2 a - b)^{0}}$. Recall $b^0 = 1$ and anything to zero power except zero is 1: $$(2 a - b^{0}) = (2 a - 1)$$ $$(2 a - b)^{0} = 1$$ Thus the expression is: $$\frac{2 a - 1}{1} = 2 a - 1$$ 29. Simplify $\frac{3 x^{0} - n^{-2}}{-(3 m - n)}$. Rewrite $3 x^0 = 3$, $n^{-2} = \frac{1}{n^{2}}$: $$\frac{3 - \frac{1}{n^{2}}}{-(3 m - n)} = \frac{\frac{3 n^{2} - 1}{n^{2}}}{-(3 m - n)} = \frac{3 n^{2} - 1}{-n^{2} (3 m - n)}$$ 30. Simplify $\frac{(5 p q r)^{-1}}{(15 p^{-1} q r^{-1})^{-1}}$. Rewrite: $$(5 p q r)^{-1} = \frac{1}{5 p q r}$$ $$(15 p^{-1} q r^{-1})^{-1} = \frac{1}{15 p^{-1} q r^{-1}}^{-1} = 15 p^{-1} q r^{-1} = 15 \cdot \frac{1}{p} \cdot q \cdot \frac{1}{r} = \frac{15 q}{p r}$$ So the original expression: $$\frac{\frac{1}{5 p q r}}{\frac{15 q}{p r}} = \frac{1}{5 p q r} \cdot \frac{p r}{15 q} = \frac{1}{75 q^2}$$ 31. Simplify $\frac{2^{10} \cdot 3^{15}}{9 \cdot 3^{10} \cdot 12}$. Rewrite $9 = 3^2$, $12 = 2^2 \cdot 3$. Substitute: $$\frac{2^{10} 3^{15}}{3^2 \cdot 3^{10} \cdot 2^2 3} = \frac{2^{10} 3^{15}}{2^{2} 3^{13}} = 2^{10 - 2} 3^{15 - 13} = 2^{8} 3^{2} = 256 \cdot 9 = 2304$$ 32. Simplify $\frac{2^{12} \cdot 5^{18}}{10^{12}}$. Rewrite $10 = 2 \cdot 5$: $$10^{12} = (2 \cdot 5)^{12} = 2^{12} 5^{12}$$ Divide: $$\frac{2^{12} 5^{18}}{2^{12} 5^{12}} = 5^{18 - 12} = 5^6 = 15625$$ 33. Simplify $\frac{24^{5}}{32 \cdot 12^{5}}$. Rewrite: $24 = 2^{3} \cdot 3$, $12 = 2^{2} \cdot 3$, $32 = 2^{5}$. Calculate: $$24^5 = (2^3 \cdot 3)^5 = 2^{15} 3^{5}$$ $$12^5 = (2^{2} \cdot 3)^5 = 2^{10} 3^{5}$$ Expression: $$\frac{2^{15} 3^{5}}{2^{5} \cdot 2^{10} 3^{5}} = \frac{2^{15} 3^{5}}{2^{15} 3^{5}} = 1$$ 34. Simplify $\frac{64 \cdot 5^{4}}{2^{8} \cdot 3^{2}}$. Rewrite $64 = 2^6$: $$\frac{2^{6} 5^{4}}{2^{8} 3^{2}} = 2^{6 - 8} 5^{4} 3^{-2} = 2^{-2} 5^{4} 3^{-2} = \frac{5^{4}}{2^{2} 3^{2}} = \frac{625}{36}$$ 35. Simplify $\left(\frac{144 \cdot 125}{2^{3} \cdot 3^{3}}\right)^{-1}$. Rewrite: $144 = 2^{4} 3^{2}$, $125 = 5^{3}$ Calculate numerator: $144 \cdot 125 = 2^{4} 3^{2} 5^{3}$ Rewrite denominator: $2^{3} 3^{3}$ Divide inside parentheses: $$\frac{2^{4} 3^{2} 5^{3}}{2^{3} 3^{3}} = 2^{4-3} 3^{2 - 3} 5^{3} = 2^{1} 3^{-1} 5^{3} = \frac{2 \cdot 125}{3} = \frac{250}{3}$$ Invert because of the negative exponent: $$\left(\frac{250}{3}\right)^{-1} = \frac{3}{250}$$ 36. Simplify $\left(\frac{72 \cdot 5^{2}}{125 \cdot 3^{2}}\right)^{-1}$. Rewrite: $72 = 2^{3} \cdot 3^{2}$, $125 = 5^{3}$ Calculate numerator: $$72 \cdot 5^{2} = 2^{3} 3^{2} 5^{2}$$ Calculate denominator: $$125 \cdot 3^{2} = 5^{3} 3^{2}$$ Divide inside: $$\frac{2^{3} 3^{2} 5^{2}}{5^{3} 3^{2}} = 2^{3} 3^{2-2} 5^{2-3} = 2^{3} 5^{-1} = \frac{8}{5}$$ Invert final result: $$\left(\frac{8}{5}\right)^{-1} = \frac{5}{8}$$