1. Simplify
a. (2-1 - 3-1)-1 + (6-1 - 8-1)-1
\begin{align*} (2^{-1} - 3^{-1})^{-1} + (6^{-1} - 8^{-1})^{-1} \end{align*}
Solution:
\begin{align*} \text{Formula } \\ & \boxed{x^{-1} = \frac{1}{x}} \quad \boxed{\left(\frac{p}{q}\right)^{-m} = \left(\frac{q}{p}\right)^{m}} \\ \\ &= \left(\frac{1}{2} - \frac{1}{3}\right)^{-1} + \left(\frac{1}{6} - \frac{1}{8}\right)^{-1} \\ \\ &= \left(\frac{3}{6} - \frac{2}{6}\right)^{-1} + \left(\frac{4}{24} - \frac{3}{24}\right)^{-1} \\ \\ &= \left(\frac{1}{6}\right)^{-1} + \left(\frac{1}{24}\right)^{-1} \\ \\ &= 6 + 24 \\ \\ &= 30 \end{align*}
b. (2-1 x 3-1)2 x (-3/8)-1
\begin{align*} (2^{-1} \times 3^{-1})^{2} \times \left(-\frac{3}{8}\right)^{-1} \\ \end{align*}
Solution:
\begin{align*} \text{Formula } \\ & \boxed{x^{-1} = \frac{1}{x}} \quad \boxed{\left(\frac{p}{q}\right)^{-m} = \left(\frac{q}{p}\right)^{m}} \\ \\ &= \left(\frac{1}{2} \times \frac{1}{3}\right)^{2} \times \left(-\frac{8}{3}\right) \\ \\ &= \left(\frac{1}{6}\right)^{2} \times \left(-\frac{8}{3}\right) \\ \\ &= \left(\frac{1}{\cancelto{9}{36}}\right) \times \left(-\frac{\cancelto{2}8}{3}\right) \\ \\ &= -\frac{2}{27} \end{align*}
c. (4-1 x 3-1) ÷ (12)-1
\begin{align*} (4^{-1} \times 3^{-1}) \div 12^{-1} \\ \end{align*}
Solution:
\begin{align*} \text{Formula } \\ & \boxed{x^{-1} = \frac{1}{x}} \\ \\ &= (4^{-1} \times 3^{-1}) \div 12^{-1} \\ \\ &= \left(\frac{1}{4} \times \frac{1}{3}\right) \div \frac{1}{12} \\ \\ &= \left(\frac{1}{12}\right) \div \frac{1}{12} \\ \\ &= \frac{1}{\cancel{12}} \times \cancel{12} \\ \\ &= 1 \end{align*}
d. (1/2)-2 + (1/3)-2 + (1/4)-2
\begin{align*} \left(\frac{1}{2}\right)^{-2} + \left(\frac{1}{3}\right)^{-2} + \left(\frac{1}{4}\right)^{-2} \\ \\ \end{align*}
Solution:
\begin{align*} \text{Formula } \\ & \boxed{\left(\frac{p}{q}\right)^{-m} = \left(\frac{q}{p}\right)^{m}} \\ \\ &= \left(\frac{2}{1}\right)^{2} + \left(\frac{3}{1}\right)^{2} + \left(\frac{4}{1}\right)^{2} \\ \\ & = (2)^2 + (3)^2 + (4)^2 \\ \\ &= 4 + 9 + 16 \\ \\ &= 29 \end{align*}
e. (2-1 ÷ 5-1)2 x (-5/8)-1
\begin{align*} &= \left(2^{-1} \div 5^{-1}\right)^2 \times \left(-\frac{5}{8}\right)^{-1} \\ \end{align*}
Solution:
\begin{align*} \text{Formula } \\ & \boxed{x^{-1} = \frac{1}{x}} \\ \\ \\ &= \left(\frac{1}{2} \div \frac{1}{5}\right)^2 \times \left(-\frac{5}{8}\right)^{-1} \\ \\ &= \left(\frac{1}{2} \times \frac{5}{1}\right)^2 \times \left(-\frac{5}{8}\right)^{-1} \\ \\ &= \left(\frac{5}{2}\right)^2 \times \left(-\frac{8}{5}\right) \\ \\ &= \frac{\cancelto{5}{25}}{\cancel4} \times \left(-\frac{\cancelto{2}8}{\cancel5}\right) \\ \\ &= 5 \times -2 \\ \\ &= -10 \end{align*}
f. (3/7)-2 ÷ (4/7)-2
\begin{align*} {\left(\frac{3}{7}\right)^{-2}} \div {\left(\frac{4}{7}\right)^{-2}} \\ \end{align*}
Solution:
\begin{align*} \text{Formula } \\ & \boxed{\left(\frac{p}{q}\right)^{-m} = \left(\frac{q}{p}\right)^{m}} \\ \\ &= {\left(\frac{7}{3}\right)^{2}} \div {\left(\frac{7}{4}\right)^{2}} \\ \\ &= \frac{49}{9} \div \frac{49}{16} \\ \\ &= \frac{\cancel{49}}{9} \times \frac{16}{\cancel{49}} \\ \\ &= \frac{16}{9} \end{align*}
g. [(2/5)-1 x (3/4)-1]-1
\begin{align*} \left[\left(\frac{2}{5}\right)^{-1} \times \left(\frac{3}{4}\right)^{-1}\right]^{-1} \end{align*}
Solution:
\begin{align*} \text{Formula } \\ & \boxed{\left(\frac{p}{q}\right)^{-m} = \left(\frac{q}{p}\right)^{m}} \\ \\ &= \left[\frac{5}{\cancel{2}} \times \frac{\cancelto{2}4}{3}\right]^{-1} \\ \\ &= \left[\frac{5 \times 2}{1 \times 3}\right]^{-1} \\ \\ &= \left[\frac{10}{3}\right]^{-1} \\ \\ &= \frac{3}{10} \\ \end{align*}
h. (45 ÷ 48) x 64
\begin{align*} ({4^5} \div {4^8}) \times 64 \end{align*}
Solution:
\begin{align*} \text{Formula } \\ & \boxed{x^m \div x^n = \left(\frac{1}{x^{n - m}}\right)} \quad (m < n) \\ \\ &= \left(\frac{1}{4^{8-5}}\right) \times 64 \\ \\ &= \frac{1}{4^3} \times 64 \\ \\ &= \frac{1}{64} \times 64 \\ \\ &= 1 \end{align*}
i. (4/5)-3 x (4/5)2 x (4/5)3
\begin{align*} & \left(\frac{4}{5}\right)^{-3} \times \left(\frac{4}{5}\right)^{2} \times \left(\frac{4}{5}\right)^{3} \\ \\ \end{align*}
Solution:
\begin{align*} & \text{Formula } \\ a^{m} \times a^{n} &= a^{m+n} \\ \\ & = \left(\frac{4}{5}\right)^{-3 + 2 + 3} \\ \\ & = \left(\frac{4}{5}\right)^{2} \\ \\ & = \left(\frac{4}{5} \times \frac{4}{5}\right) \\ \\ & = \frac{16}{25} \end{align*}
j. (-1/2)-3 / (-1/2)-4 - (-1/2)-5 / (-1/2)-6
\begin{align*} &= \frac{\left(-\frac{1}{2}\right)^{-3}}{\left(-\frac{1}{2}\right)^{-4}} - \frac{\left(-\frac{1}{2}\right)^{-5}}{\left(-\frac{1}{2}\right)^{-6}} \\ \\ \end{align*}
Solution:
\begin{align*} &\text{Formula } \\ \\ x^{-m} &= \frac{1}{x^m} \\ \\ x^m \div x^n &= x^{m-n} \\ \\ &= \left(-\frac{1}{2}\right)^{(-3) - (-4)} - \left(-\frac{1}{2}\right)^{(-5) - (-6)} \\ \\ &= \left(-\frac{1}{2}\right)^{-3 + 4} - \left(-\frac{1}{2}\right)^{-5 + 6} \\ \\ &= \left(-\frac{1}{2}\right)^1 - \left(-\frac{1}{2}\right)^1 \\ \\ &= -\frac{1}{2} + \frac{1}{2} \\ \\ &= 0 \end{align*}
2. By what number should we multiply (2-5) so that the product may be equal to (2-1)?
Solution:
\begin{align*} \text{Let } (2^{-5}) & \text{ be multiplied by x} \\ \\ x \times 2^{-5} &= 2^{-1} \\ \\ x &= 2^{-1} \div 2^{-5} \\ \\ x &= 2^{(-1) - (-5)} \\ \\ x &= 2^{-1 + 5} \\ \\ x &= 2^{4} \\ \\ x &= 16 \\ \end{align*}
3. By what number should (-1/5)-1 be multiplied so that the product may be equal to (1/5)-3?
Solution:
\begin{align*} \text{Let } \left(-\frac{1}{5}\right)^{-1} & \text{ be multiplied by x} \\ \\ \left(-\frac{1}{5}\right)^{-1} \times x &= \left(\frac{1}{5}\right)^{-3} \\ \\ x &= \left(\frac{1}{5}\right)^{-3} \div \left(-\frac{1}{5}\right)^{-1} \\ \\ x &= (5)^{3} \div \left(-\frac{5}{1}\right)^{1} \\ \\ x &= 125 \div \left(-\frac{5}{1}\right)^{1} \\ \\ x &= \cancelto{25}{125} \times \left(-\frac{1}{\cancel5}\right) \\ \\ x &= 25 \times -1 \\ \\ x &= -25 \\ \end{align*}
4. By what number should (24)-1 be divided so that the quotient may be equal to (4)-1?
Solution:
\begin{align*} \text{Let } (24)^{-1} & \text{ be divided by x} \\ \\ (24)^{-1} \div x &= (4)^{-1} \\ \\ \left(\frac{1}{24}\right) \times \left(\frac{1}{x}\right) &= \left(\frac{1}{4}\right) \\ \\ \left(\frac{1}{24x}\right) &= \left(\frac{1}{4}\right) \\ \\ 4 &= 24x \\ \\ \left(\frac{4}{24}\right) &= x \\ \\ \left(\frac{1}{6}\right) &= x \\ \\ x & = \left(\frac{1}{6}\right) \\ \\ \end{align*}
5. By what number should (-2/3)3 be divided so that the quotient may be equal to (9/4)-2?
Solution:
\begin{align*} \text{Let } \left(-\frac{2}{3}\right)^{3} & \text{ be divided by x} \\ \\ \left(-\frac{2}{3}\right)^{3} \div x &= \left(\frac{9}{4}\right)^{-2} \\ \\ \left(-\frac{2}{3}\right)^{3} \times \left(\frac{1}{x}\right) &= \left(\frac{9}{4}\right)^{-2} \\ \\ \left(-\frac{8}{27}\right)^{3} \times \left(\frac{1}{x}\right) &= \left(\frac{4}{9}\right)^{2} \\ \\ \left(-\frac{8}{27}\right) \times \left(\frac{1}{x}\right) &= \left(\frac{16}{81}\right) \\ \\ \left(-\frac{8}{27}\right) \times \left(\frac{81}{16}\right) &= x \\ \\ x & = \left(-\frac{\cancel8}{\cancel{27}}\right) \times \left(\frac{\cancelto{3}{81}}{\cancelto{2}{16}}\right) \\ \\ x &= -\frac{3}{2} \end{align*}
6. Find the value of x so that -
a. (3/4)-9 x (3/4)-7 = (3/4)4x
\begin{align*} \left(\frac{3}{4}\right)^{-9} \times \left(\frac{3}{4}\right)^{-7} &= \left(\frac{3}{4}\right)^{4x} \\ \end{align*}
Solution:
\begin{align*} \left(\frac{3}{4}\right)^{-9 + (-7)} &= \left(\frac{3}{4}\right)^{4x} \\ \\ \left(\frac{3}{4}\right)^{-9 -7} &= \left(\frac{3}{4}\right)^{4x} \\ \\ \left(\frac{3}{4}\right)^{-16} &= \left(\frac{3}{4}\right)^{4x} \\ \\ \text{Since bases are same, } & \text{their exponents must be equal.} \\ \\ 4x &= -16 \\ \\ x &= \frac{-16}{4} \\ \\ x &= -4 \end{align*}
b. (2/9)-6 x (2/9)3 = (2/9)2x - 1
\begin{align*} \left(\frac{2}{9}\right)^{-6} \times \left(\frac{2}{9}\right)^{3} &= \left(\frac{2}{9}\right)^{2x - 1} \\ \end{align*}
Solution:
\begin{align*} \left(\frac{2}{9}\right)^{-6 + 3} &= \left(\frac{2}{9}\right)^{2x - 1} \\ \\ \left(\frac{2}{9}\right)^{-3} &= \left(\frac{2}{9}\right)^{2x - 1} \\ \\ \text{Since bases are same, } & \text{their exponents must be equal.} \\ \\ 2x - 1 &= -3 \\ \\ 2x &= -3 + 1 \\ \\ 2x &= -2 \\ \\ x &= \frac{-2}{2} \\ \\ x &= -1 \end{align*}
c. (5/7)2 ÷ (5/7)3x + 1 = (5/7)4
\begin{align*} \left(\frac{5}{7}\right)^2 \div \left(\frac{5}{7}\right)^{3x + 1} &= \left(\frac{5}{7}\right)^{4} \\ \end{align*}
Solution:
\begin{align*} \left(\frac{5}{7}\right)^{2 - (3x + 1)} & = \left(\frac{5}{7}\right)^{4} \\ \\ \left(\frac{5}{7}\right)^{2 - 3x - 1} & = \left(\frac{5}{7}\right)^{4} \\ \\ \left(\frac{5}{7}\right)^{1 - 3x} & = \left(\frac{5}{7}\right)^{4} \\ \\ \text{Since bases are same, } & \text{their exponents must be equal.} \\ \\ 1 - 3x &= 4 \\ \\ 3x &= 4 - 1 \\ \\ -3x &= 3 \\ \\ x &= \frac{3}{-3} \\ \\ x &= -1 \end{align*}
d. (-6/11)x ÷ [(-6/11)-2]-1 = [(-6/11)2]-3
\begin{align*} \left(-\frac{6}{11}\right)^x \div \left[\left(-\frac{6}{11}\right)^{-2}\right]^{-1} &= \left[\left(-\frac{6}{11}\right)^2\right]^{-3} \\ \end{align*}
Solution:
\begin{align*} \left(-\frac{6}{11}\right)^x \div \left(-\frac{6}{11}\right)^2 &= \left(-\frac{6}{11}\right)^{-6} \\ \\ \left(-\frac{6}{11}\right)^{x-2} &= \left(-\frac{6}{11}\right)^{-6} \\ \\ \text{Since bases are same, } & \text{their exponents must be equal.} \\ \\ x - 2 &= -6 \\ \\ x &= -6 + 2 \\ \\ x &= -4 \end{align*}
7. If p/q = (2/3)2 x (1/3)-4 , find the value of (p/q)-2
\begin{align*} \frac{p}{q} &= \left(\frac{2}{3}\right)^2 \times \left(\frac{1}{3}\right)^{-4} \quad \text{Find } \left(\frac{p}{q}\right)^{-2} \\ \end{align*}
Solution:
\begin{align*} \frac{p}{q} &= \left(\frac{2}{3}\right)^2 \times (3)^{4} \\ \\ &= \frac{2}{\cancel3} \times \frac{2}{\cancel3} \times 3 \times 3 \times \cancel3 \times \cancel3 \\ \\ &= 2 \times 2 \times \times 3 \times 3 \\ \\ \frac{p}{q} &= 36 \\ \\ \left(\frac{p}{q}\right)^{-2} &= \left(36\right)^{-2} \\ \\ &= \left(\frac{1}{36}\right)^2 \\ \\ &= \left(\frac{1}{36}\right) \times \left(\frac{1}{36}\right) \\ \\ \left(\frac{p}{q}\right)^{-2} &= \frac{1}{1296} \end{align*}
8. If a = (3/5)-2 ÷ (7/5)0 , find the value of a-3
\begin{align*} a &= \left(\frac{3}{5}\right)^{-2} \div \left(\frac{7}{5}\right)^0 \quad \text{Find } a^{-3} \\ \end{align*}
Solution:
\begin{align*} a &= \left(\frac{3}{5}\right)^{-2} \div \left(\frac{7}{5}\right)^0 \\ \\ a &= \left(\frac{5}{3}\right)^{2} \div 1 \\ \\ &= \left(\frac{5}{3}\right)^2 \\ \\ &= \frac{25}{9} \\ \\ a^{-3} &= \left(\frac{25}{9}\right)^{-3} \\ \\ &= \left(\frac{9}{25}\right)^3 \\ \\ &= \left(\frac{9}{25}\right) \times \left(\frac{9}{25}\right) \times \left(\frac{9}{25}\right) \\ \\ a^{-3} &= \frac{729}{15625} \end{align*}
9. Simplify [(2/3)2]3 x (2/3)-4 x 3-1 x 1/6
\begin{align*} &\left[\left(\frac{2}{3}\right)^2\right]^3 \times \left(\frac{2}{3}\right)^{-4} \times 3^{-1} \times \frac{1}{6} \\ \end{align*}
Solution:
\begin{align*} &= \left(\frac{2}{3}\right)^{6} \times \left(\frac{2}{3}\right)^{-4} \times \frac{1}{3} \times \frac{1}{6} \\ \\ &= \left(\frac{2}{3}\right)^{6 + (-4)} \times \frac{1}{3} \times \frac{1}{6} \\ \\ &= \left(\frac{2}{3}\right)^{(6 -4)} \times \frac{1}{18} \\ \\ &= \left(\frac{2}{3}\right)^{2} \times \frac{1}{18} \\ \\ &= \frac{\cancelto{2}{4}}{9} \times \frac{1}{\cancelto{9}{18}} \\ \\ &= \frac{2}{9 \times 9} \\ \\ &= \frac{2}{81} \end{align*}
10. Find the reciprocals of
a. (1/2)-2 ÷ (2/3)-3
\begin{align*} {\left(\frac{1}{2}\right)^{-2}} \div {\left(\frac{2}{3}\right)^{-3}} \\ \\ \end{align*}
Solution:
\begin{align*} &= 2^2 \div {\left(\frac{3}{2}\right)^3} \\ \\ &= 2^2 \div {\left(\frac{3}{2} \times \frac{3}{2} \times \frac{3}{2}\right)} \\ \\ &= 4 \div \frac{27}{8} \\ \\ &= 4 \times \frac{8}{27} \\ \\ &= \frac{32}{27} \\ \\ \text{Reciprocal } &= \frac{27}{32}. \end{align*}
b. (2/5)3 x (5/4)2
\begin{align*} \left(\frac{2}{5}\right)^3 \times \left(\frac{5}{4}\right)^2 \end{align*}
Solution:
\begin{align*} & = \frac{2}{5} \times \frac{\cancel2}{\cancel5} \times \frac{\cancel2}{\cancel5} \times \frac{\cancel5}{\cancelto{2}4} \times \frac{\cancel5}{\cancelto{2}4} \\ \\ &= \frac{2}{20} \\ \\ &= \frac{1}{10} \\ \\ \text{Reciprocal } &= {10} \end{align*}
11. Express the following as a rational number with positive exponent.
a. (3/2)-4
\begin{align*} \left(\frac{3}{2}\right)^{-4} \end{align*}
Solution:
\begin{align*} &= \left(\frac{2}{3}\right)^{4} \\ \end{align*}
b. (3-3)2
\begin{align*} (3^{-3})^2 \end{align*}
Solution:
\begin{align*} &= 3^{-3 \times 2} \\ \\ &= 3^{-6} \\ \\ &= \frac{1}{3^6} \\ \\ \end{align*}
c. 72 x 7-3
\begin{align*} 7^2 \times 7^{-3} \end{align*}
Solution:
\begin{align*} &= 7^{2+ (- 3)} \\ \\ &= 7^{2 - 3} \\ \\ &= 7^{-1} \\ \\ &= \frac{1}{7} \end{align*}
d. [(5/8)-2]3
\begin{align*} \left(\left(\frac{5}{8}\right)^{-2}\right)^3 \end{align*}
Solution:
\begin{align*} &= \left(\frac{5}{8}\right)^{-2 \times 3} \\ &= \left(\frac{5}{8}\right)^{-6} \\ &= \left(\frac{8}{5}\right)^{6} \end{align*}
e. (2/5)-3 x (2/5)5
\begin{align*} \left(\frac{2}{5}\right)^{-3} \times \left(\frac{2}{5}\right)^{5} \end{align*}
Solution:
\begin{align*} &= \left(\frac{2}{5}\right)^{-3 + 5} \\ &= \left(\frac{2}{5}\right)^{2} \end{align*}
f. (83 ÷ 85) x 8-4
\begin{align*} (8^{3} \div 8^{5}) \times 8^{-4} \end{align*}
Solution:
\begin{align*} &= 8^{3 - 5} \times 8^{-4} \\ \\ &= 8^{-2} \times 8^{-4} \\ \\ &= 8^{-2 + (-4)} \\ \\ &= 8^{-2 -4} \\ \\ &= 8^{-6} \\ \\ &= \left(\frac{1}{8}\right)^{6} \end{align*}
12. Express the following as a rational number with negative exponent.
a. (1/7)5
\begin{align*} \left(\frac{1}{7}\right)^{5} \end{align*}
Solution:
\begin{align*} &= \left(\frac{7}{1}\right)^{5} \\ \\ &= (7)^{-5} \end{align*}
b. (32)9
\begin{align*} (3^2)^9 \end{align*}
Solution:
\begin{align*} & = (3)^{2 \times 9} \\ \\ & = (3)^{18} \\ \\ &= \left(\frac{1}{3}\right)^{-18} \\ \end{align*}
c. 53 x 52
\begin{align*} 5^3 \times 5^2 \end{align*}
Solution:
\begin{align*} & = 5^{3 + 2} \\ \\ & = 5^{5} \\ \\ &= \left(\frac{1}{5}\right)^{-5} \\ \end{align*}
d. ((-8/9)3)2
\begin{align*} \left(\left(-\frac{8}{9}\right)^3\right)^2 \end{align*}
Solution:
\begin{align*} &= \left(-\frac{8}{9}\right)^{3 \times 2} \\ \\ &= \left(-\frac{8}{9}\right)^{6} \\ \\ &= \left(-\frac{9}{8}\right)^{-6} \end{align*}
e. (5/7)3 ÷ (5/7)2
\begin{align*} \left(\frac{5}{7}\right)^3 \div \left(\frac{5}{7}\right)^2 \end{align*}
Solution:
\begin{align*} &= \left(\frac{5}{7}\right)^{3 - 2} \\ \\ &= \left(\frac{5}{7}\right)^{1} \\ \\ &= \left(\frac{7}{5}\right)^{-1} \end{align*}
f. (26 ÷ 25) x 22
\begin{align*} &= \left(2^6 \div 2^5\right) \times 2^2 \\ \end{align*}
Solution:
\begin{align*} &= (2)^{(6-5)} \times 2^2 \\ \\ &= (2)^1 \times 2^2 \\ \\ &= (2)^{1+2} \\ \\ &= (2)^3 \\ \\ &= \left(\frac{1}{2}\right)^{-3} \end{align*}