CHAPTER 10 - EXPONENTS AND POWERS

Using the negative exponent rule: \(a^{-n} = \frac{1}{a^n}\)

\[ 3^{-2} = \frac{1}{3^2} = \frac{1}{9} \]

Using the negative exponent rule: \(a^{-n} = \frac{1}{a^n}\)

\[ (-4)^{-2} = \frac{1}{(-4)^2} = \frac{1}{16} \]

Using the negative exponent rule: \(\left(\frac{a}{b}\right)^{-n} = \left(\frac{b}{a}\right)^n\)

\[ \left(\frac{1}{2}\right)^{-5} = \left(\frac{2}{1}\right)^5 = 2^5 = 32 \]

Using the quotient of powers rule: \(a^m \div a^n = a^{m-n}\)

\[ (-4)^5 \div (-4)^8 = (-4)^{5-8} = (-4)^{-3} \]

Convert to positive exponent:

\[ (-4)^{-3} = \frac{1}{(-4)^3} = \frac{1}{-64} = -\frac{1}{64} \]

Using the power of a power rule: \((a^m)^n = a^{m \times n}\)

\[ \left(\frac{1}{2^3}\right)^2 = \frac{1^2}{(2^3)^2} = \frac{1}{2^{3 \times 2}} = \frac{1}{2^6} = \frac{1}{64} \]

Using the power of a product rule: \(a^n \times b^n = (a \times b)^n\)

\[ (-3)^4 \times \left(\frac{5}{3}\right)^4 = \left((-3) \times \frac{5}{3}\right)^4 = (-5)^4 = 625 \]

First, simplify the division using quotient rule: \(a^m \div a^n = a^{m-n}\)

\[ 3^{-7} \div 3^{-10} = 3^{-7 - (-10)} = 3^{-7 + 10} = 3^3 \]

Now multiply: \(3^3 \times 3^{-5} = 3^{3 + (-5)} = 3^{-2}\)

Convert to positive exponent:

\[ 3^{-2} = \frac{1}{3^2} = \frac{1}{9} \]

Using the power of a product rule: \(a^n \times b^n = (a \times b)^n\)

\[ 2^{-3} \times (-7)^{-3} = (2 \times (-7))^{-3} = (-14)^{-3} \]

Convert to positive exponent:

\[ (-14)^{-3} = \frac{1}{(-14)^3} = \frac{1}{-2744} = -\frac{1}{2744} \]

Calculate each part separately:

\[ 3^0 = 1 \quad (\text{any number to the power 0 is 1}) \]

\[ 4^{-1} = \frac{1}{4} \]

\[ 2^2 = 4 \]

Now combine:

\[ (1 + \frac{1}{4}) \times 4 = \left(\frac{5}{4}\right) \times 4 = 5 \]

Calculate each part separately:

\[ 2^{-1} = \frac{1}{2}, \quad 4^{-1} = \frac{1}{4} \]

\[ 2^{-1} \times 4^{-1} = \frac{1}{2} \times \frac{1}{4} = \frac{1}{8} \]

\[ 2^{-2} = \frac{1}{2^2} = \frac{1}{4} \]

Now divide:

\[ \frac{1}{8} \div \frac{1}{4} = \frac{1}{8} \times \frac{4}{1} = \frac{4}{8} = \frac{1}{2} \]

Convert negative exponents to positive:

\[ \left(\frac{1}{2}\right)^{-2} = 2^2 = 4 \]

\[ \left(\frac{1}{3}\right)^{-2} = 3^2 = 9 \]

\[ \left(\frac{1}{4}\right)^{-2} = 4^2 = 16 \]

Now add:

\[ 4 + 9 + 16 = 29 \]

Any expression to the power of 0 is equal to 1:

\[ (3^{-1} + 4^{-1} + 5^{-1})^0 = 1 \]

First, simplify the inner expression:

\[ \left(\frac{-2}{3}\right)^{-2} = \left(\frac{3}{-2}\right)^2 = \frac{9}{4} \]

Now square the result:

\[ \left(\frac{9}{4}\right)^2 = \frac{81}{16} \]

Convert negative exponents to positive:

\[ 8^{-1} = \frac{1}{8}, \quad 2^{-4} = \frac{1}{2^4} = \frac{1}{16} \]

Now substitute:

\[ \frac{\frac{1}{8} \times 125}{\frac{1}{16}} = \frac{125}{8} \times \frac{16}{1} = \frac{125 \times 16}{8} = \frac{125 \times 2}{1} = 250 \]

Convert negative exponents to positive:

\[ 5^{-1} = \frac{1}{5}, \quad 2^{-1} = \frac{1}{2}, \quad 6^{-1} = \frac{1}{6} \]

Now multiply:

\[ \left(\frac{1}{5} \times \frac{1}{2}\right) \times \frac{1}{6} = \frac{1}{10} \times \frac{1}{6} = \frac{1}{60} \]

First, simplify the expressions inside the parentheses:

\[ \left(\frac{1}{3}\right)^{-1} = 3, \quad \left(\frac{1}{4}\right)^{-1} = 4 \]

Now subtract:

\[ 3 - 4 = -1 \]

Now apply the outer exponent:

\[ (-1)^{-1} = \frac{1}{-1} = -1 \]

Convert negative exponents to positive:

\[ \left(\frac{5}{8}\right)^{-7} = \left(\frac{8}{5}\right)^7, \quad \left(\frac{8}{5}\right)^{-4} = \left(\frac{5}{8}\right)^4 \]

Now multiply:

\[ \left(\frac{8}{5}\right)^7 \times \left(\frac{5}{8}\right)^4 = \left(\frac{8}{5}\right)^7 \times \left(\frac{8}{5}\right)^{-4} = \left(\frac{8}{5}\right)^{7 - 4} = \left(\frac{8}{5}\right)^3 = \frac{512}{125} \]

First, express all numbers as powers of 5:

\[ 25 = 5^2, \quad 10 = 2 \times 5, \quad 5^{-3} = \frac{1}{5^3} \]

Now substitute:

\[ \frac{5^2 \times t^{-4}}{\frac{1}{5^3} \times 2 \times 5 \times t^{-8}} = \frac{5^2 \times t^{-4}}{\frac{2 \times 5}{5^3} \times t^{-8}} = \frac{5^2 \times t^{-4} \times 5^3}{2 \times 5 \times t^{-8}} \]

Simplify:

\[ = \frac{5^{2+3} \times t^{-4}}{2 \times 5 \times t^{-8}} = \frac{5^5 \times t^{-4}}{2 \times 5 \times t^{-8}} = \frac{5^{5-1} \times t^{-4 - (-8)}}{2} = \frac{5^4 \times t^{4}}{2} = \frac{625 t^4}{2} \]

Express numbers as powers of prime factors:

\[ 10 = 2 \times 5, \quad 125 = 5^3, \quad 6 = 2 \times 3 \]

Now substitute:

\[ \frac{3^{-5} \times (2 \times 5)^{-5} \times 5^3}{5^{-7} \times (2 \times 3)^{-5}} = \frac{3^{-5} \times 2^{-5} \times 5^{-5} \times 5^3}{5^{-7} \times 2^{-5} \times 3^{-5}} \]

Cancel common terms:

\[ = \frac{5^{-5} \times 5^3}{5^{-7}} = 5^{-5 + 3 - (-7)} = 5^{-2 + 7} = 5^5 = 3125 \]