Lesson 3: Exponential Notation
- 1. 𝐏𝐓𝐒 𝟑 Bridge to Calculus Workshop Summer 2020 Lesson 3 Exponential Notation "In mathematics the art of proposing a question must be held of higher value than solving it." – Georg Cantor -
- 2. Lehman College, Department of Mathematics Integer Exponents (1 of 4) A product of identical numbers is usually written in exponential notation. For example: DEFINITION. Let 𝑎 be a real number and let 𝑛 ≥ 2 be a positive integer, then the 𝑛-th power of 𝑎 is defined as: The number 𝑎 is called the base, and 𝑛 is the exponent. Example 1. Determine the following: 𝟓 ⋅ 𝟓 ⋅ 𝟓 = 𝟓 𝟑 𝒂 𝒏 = 𝒂 ⋅ 𝒂 ⋅ ⋯ ⋅ 𝒂 𝒏 𝐟𝐚𝐜𝐭𝐨𝐫𝒔 1 2 4 (a) (b) −2 3 (c) −23
- 3. Lehman College, Department of Mathematics Integer Exponents (2 of 4) Solution: Note the difference between (b) and (c): Let us discover some of the rules for working with exponential notation: It appears that when we multiply two powers of the same base, we add their exponents. 1 4 3 =(a) (b) −2 4 = (c) −24 = 1 4 1 4 1 4 = 1 64 −2 ⋅ −2 ⋅ −2 ⋅ −2 = 16 − 2 ⋅ 2 ⋅ 2 ⋅ 2 = −16 53 ⋅ 54 = 5 ⋅ 5 ⋅ 5 ⋅ 3 factors 5 ⋅ 5 ⋅ 5 ⋅ 5 = 4 factors 57
- 4. Lehman College, Department of Mathematics Integer Exponents (3 of 4) In general, for any real number 𝑎 and positive integers 𝑛 and 𝑚 with 𝑛, 𝑚 ≥ 2: Let us look at division of powers. Consider: It appears that when we find the quotient of two powers of the same base, we subtract their exponents. Then, for any real number 𝑎 (𝑎 ≠ 0) and any positive integers 𝑛 and 𝑚 with 𝑛, 𝑚 ≥ 2 and 𝑛 − 𝑚 ≥ 2, then: 𝒂 𝒎 𝒂 𝒏 = 𝒂 𝒎+𝒏 55 53 = 5 ⋅ 5 ⋅ 5 ⋅ 5 ⋅ 5 5 ⋅ 5 ⋅ 5 = 5 ⋅ 5 = 52 𝒂 𝒎 𝒂 𝒏 = 𝒂 𝒎−𝒏
- 5. Lehman College, Department of Mathematics Integer Exponents (3 of 4) The formulas we have discovered so far are true for exponents 𝑛 with 𝑛 ≥ 2. Consider the following example If we want the formula to still work, then we require that: Similarly, consider the following example: If we want the formula to still work, then we require that: 5 53−2 = 53 52 = 5 ⋅ 5 ⋅ 5 5 ⋅ 5 = 515 = 1 53 53 = 5 ⋅ 5 ⋅ 5 5 ⋅ 5 ⋅ 5 = 53−3 = 501 =
- 6. Lehman College, Department of Mathematics Integer Exponents (3 of 4) In general, for any nonzero real number 𝑎: Let us look at negative exponents. Consider: If we want the formula to still work, then we require that: In general, for any nonzero real number 𝑎, and for any integer 𝑛, we have: 𝒂 𝟏 = 𝒂 53 55 = 5 ⋅ 5 ⋅ 5 5 ⋅ 5 ⋅ 5 ⋅ 5 ⋅ 5 = 1 5 ⋅ 5 = 1 52 𝒂−𝒏 = 𝟏 𝒂 𝒏 𝒂 𝟎 = 𝟏and 1 52 = 53−5 = 5−2 𝒂 𝒏 = 𝟏 𝒂−𝒏 and
- 7. Lehman College, Department of Mathematics Integer Exponents (1 of 4) Example 2. Determine the following: Solution. Example 3. In Calculus I and II, we will solve problems such as the following. Our second step should be: 4 7 0 (a) (b) −3 −2 (c) 𝑥−1 4 7 0 =(a) 1 (b) −3 −2 = 1 −3 2 = 1 9 (c) 𝑥−1 = 1 𝑥1 = 1 𝑥 (d) 1 𝑥−3 (d) 1 𝑥−3 = 𝑥−(−3) = 𝑥3 𝑑 𝑑𝑥 1 𝑥5 = 𝑑 𝑑𝑥 𝑥−5(a) 1 2 1 𝑥3 𝑑𝑥 =(b) 1 2 𝑥−3 𝑑𝑥
- 8. Lehman College, Department of Mathematics Helix Nebula in Aquarius (1 of 1)
- 9. Lehman College, Department of Mathematics Integer Exponents (2 of 4) Let us discover more rules for working with exponential notation: It appears that when we raise an exponential expression to a power, we multiply the exponents. In general, for any nonzero real number 𝑎 (base) and integers 𝑛 and 𝑚 (exponents): 52 3 = 52 ⋅ 52 ⋅ 52 = 3 factors 5 ⋅ 5 5 ⋅ 5 5 ⋅ 5 = 6 factors 56 𝒂 𝒎 𝒏 = 𝒂 𝒏 𝒎 = 𝒂 𝒎𝒏 53 2 = 53 ⋅ 53 = 2 factors 5 ⋅ 5 ⋅ 5 5 ⋅ 5 ⋅ 5 = 6 factors 56
- 10. Lehman College, Department of Mathematics Integer Exponents (2 of 4) We will now look at the product and quotient of powers of different bases, but the same exponent: In general, for any nonzero real numbers 𝑎 and 𝑏 (base) and integer 𝑛 (exponent): We can similarly show that: 23 ⋅ 53 = = 2 ⋅ 5 2 ⋅ 5 2 ⋅ 5 = 3 factors 2 ⋅ 5 3 𝒂 𝒏 𝒃 𝒏 = 𝒂𝒃 𝒏 2 ⋅ 2 ⋅ 2 ⋅ 3 factors 5 ⋅ 5 ⋅ 5 = 3 factors 𝒂 𝒃 𝒏 = 𝒂 𝒏 𝒃 𝒏
- 11. Lehman College, Department of Mathematics Integer Exponents (1 of 4) Example 3. Simplify the following expressions using positive exponents only: Solution. Group constants and factors with same base: 4𝑥2 𝑦 −3 (a) (b) −35𝑥2 𝑦4 5𝑥6 𝑦−8 (c)(−7𝑥𝑦4)(−2𝑥5 𝑦6) (a) −7𝑥𝑦4 −2𝑥5 𝑦6 = −7 −2)(𝑥 ⋅ 𝑥5 )(𝑦4 𝑦6 = 14𝑥6 𝑦10 𝑦 4𝑥2 3 =(b) 4𝑥2 𝑦 −3 = 𝑦3 4𝑥2 3 = 𝑦3 43 𝑥2 3 = 𝑦3 64𝑥6
- 12. Lehman College, Department of Mathematics Integer Exponents (1 of 4) Solution (cont’d). Group factors with same base: −35𝑥2 𝑦4 5𝑥6 𝑦−8 =(c) −35 5 𝑥2 𝑥6 𝑦4 𝑦−8 = −7 𝑥2−6 𝑦4−(−8) = − 7𝑦12 𝑥4 = −7𝑥−4 𝑦12
- 13. Lehman College, Department of Mathematics Integer Exponents (1 of 4) Example 4. Simplify the following expression using positive exponents only: Solution 1. Raise factors to the required exponents, then group constants and factors with the same base: −4𝑎−2 𝑏3 −3 8𝑎𝑏 2 −4𝑎−2 𝑏3 −3 8𝑎𝑏 2 = −4 −3 𝑎−2 −3 𝑏3 −3 ⋅ 82 𝑎2 𝑏2 = 1 −4 3 ⋅ 𝑎−2 −3 𝑏3(−3) ⋅ 64𝑎2 𝑏2 = 64 −64 ⋅ 𝑎6 𝑎2 𝑏−9 𝑏2 = − 𝑎8 𝑏7 = −𝑎8 𝑏−7
- 14. Lehman College, Department of Mathematics Integer Exponents (1 of 4) Example 4. Simplify the following expression using positive exponents only: Solution 2. Raise factors to the required exponents, then group constants and factors with the same base: −4𝑎−2 𝑏3 −3 8𝑎𝑏 2 −4𝑎−2 𝑏3 −3 8𝑎𝑏 2 = 1 −4𝑎−2 𝑏3 3 ⋅ 64𝑎2 𝑏2 = 64𝑎2 𝑏2 −4 3 𝑎−2 3 𝑏3 3 = 64𝑎2 𝑏2 −64𝑎−6 𝑏9 = − 𝑎2− −6 𝑏9−2 = − 𝑎8 𝑏7
- 15. Lehman College, Department of Mathematics Incorrect Correct Common Errors to Avoid (4 of 9) 𝑏3 ⋅ 𝑏4 = 𝑏12 32 ⋅ 34 = 96 𝑏3 ⋅ 𝑏4 = 𝑏7 32 ⋅ 34 = 36 516 54 = 54 516 54 = 58 4𝑎 3 = 64𝑎3 4𝑎 3 = 4𝑎3 𝑏−𝑛 = 𝑏−𝑛 = 1 𝑏 𝑛 𝑎 + 𝑏 −1 = 1 𝑎 + 1 𝑏 𝑎 + 𝑏 −1 = 1 𝑎 + 𝑏 − 1 𝑏 𝑛