2.3 The Other Face of Powers

So far exponents have made things grow. Let us now turn them around and shrink things.

Picture a strip 32 units long. Rub out half and 16 units survive. In powers of 2, since 32 = 2^5 and 16 = 2^4, halving meant dividing by 2 = 2^1, and the exponent dipped by 1:

2^5 \div 2^1 = 2^4, \qquad 2^5 \div 2^2 = 2^3, \qquad 2^5 \div 2^3 = 2^2.

Each division subtracts the exponents.

Law of Exponents: Division n^a \div n^b = n^{a-b}, \qquad n \neq 0,\ a > b. For example 2^{90} \div 2^{30} = 2^{60}.

When the power drops to zero

What ought 2^0 to mean? The division rule settles it for us. Since 2^5 \div 2^5 is plainly 1 (any non-zero number over itself), and the rule says 2^5 \div 2^5 = 2^{5-5} = 2^0, we are forced to accept

2^0 = 1.

The Zero Exponent For any non-zero base x, \ x^a \div x^a = x^{a-a} = x^0 = 1. So anything (non-zero) raised to the power 0 equals 1. (Why insist n\ne 0? Because 0^0 is left undefined.)

When the power turns negative

Halve a 32-unit (2^5) strip seven times and you slip past 1 down to \tfrac{1}{4} unit:

2^5 \div 2^7 = 2^{5-7} = 2^{-2}, \qquad \text{and working it out, this equals } \tfrac{1}{4}.

So 2^{-2} = \dfrac{1}{4}. Halving twelve times gives 2^5 \div 2^{12} = 2^{-7}, and expanding reveals this to be \dfrac{1}{2^7} = \dfrac{1}{128}.

Negative Exponents A negative exponent signals reciprocal: n^{-a} = \frac{1}{n^a}, \qquad \text{and equivalently } \frac{1}{n^{-a}} = n^{a}, \qquad n \neq 0. For example 10^{-3} = \dfrac{1}{10^3} = \dfrac{1}{1000} and 7^{-2} = \dfrac{1}{49}.

With these meanings in place, every law of exponents keeps holding even when a and b are any integers, positive, zero, or negative:

n^a \times n^b = n^{a+b}, \qquad (n^a)^b = n^{ab}, \qquad n^a \div n^b = n^{a-b}.

Mental-Math Shortcut: Sliding the decimal Because each power of 10 adds one zero, you can multiply or divide by powers of 10 just by sliding the decimal point. Multiplying by 10^a shifts it a places right; multiplying by 10^{-a} (the same as dividing by 10^a) shifts it a places left. 3.4 \times 10^3 = 3400, \qquad 3.4 \times 10^{-2} = 0.034. No long multiplication needed, just count the jumps. This is the engine behind scientific notation in the next section.

Try This: Power Lines Write the powers of 3 down a vertical line: \dots, 3^{-2}=\tfrac{1}{9},\ 3^{-1}=\tfrac13,\ 3^0=1,\ 3^1=3,\ 3^2=9,\ 3^3=27,\ 3^4=81,\ 3^5=243,\dots,\ 3^8 = 6561. Stepping up one rung multiplies by 3; stepping down divides by 3. Use it to answer: how many times larger than 3^{-2} is 3^2? (Answer: 3^2 \div 3^{-2} = 3^4 = 81 times.) Build the same ladder for powers of 7 and read off products and quotients at a glance.

Figure it Out: Laws of Exponents

Practice

  1. Write each with positive exponents:   (i) 3^{-4}   (ii) 10^{-6}   (iii) (-8)^{-2}   (iv) (-4)^{-3}   (v) 10^{-90}
  2. Simplify and write in exponential form:   (i) 3^{-3}\times 3^7   (ii) 2^3\times 2^{-5}\times 2^6   (iii) q^4\times q^{-9}   (iv) 3^2 \times (-9)^{-1}   (v) 8^a \times 8^b
  3. Compute using n^a\times n^b = n^{a+b} (give one way each):   (i) 3^8   (ii) 2^{11}   (iii) 5^6
  4. Write each as a power of a power in at least two different ways:   (i) 9^6   (ii) 7^{12}   (iii) 4^{15}   (iv) 6^8
    1. \dfrac{1}{3^4}   (ii) \dfrac{1}{10^6}   (iii) \dfrac{1}{(-8)^2}   (iv) \dfrac{1}{(-4)^3}   (v) \dfrac{1}{10^{90}}.
    1. 3^{-3+7} = 3^4   (ii) 2^{3-5+6} = 2^4   (iii) q^{4-9} = q^{-5}   (iv) 3^2 \times \dfrac{1}{(-9)} = \dfrac{9}{-9} = -1   (v) 8^{a+b}.
    1. 3^8 = 3^4\times 3^4 = 81\times 81 = 6561   (ii) 2^{11} = 2^5\times 2^6 = 32\times 64 = 2048   (iii) 5^6 = 5^3\times 5^3 = 125\times 125 = 15625.
    1. 9^6 = (9^2)^3 = (9^3)^2; also 9^6 = (3^2)^6 = 3^{12}.   (ii) 7^{12} = (7^3)^4 = (7^4)^3.   (iii) 4^{15} = (4^3)^5 = (4^5)^3; also 4^{15} = (2^2)^{15} = 2^{30}.   (iv) 6^8 = (6^2)^4 = (6^4)^2.