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Lesson 2: Negative exponents, product and power, theorems for exponents, circle relationships We define 2 to the third power as follows: 3 2=2x2x2 We have agreed that 2 means 2 times 2 times 2. 3 In a similar fashion, we define 2 to the negative third power to mean 1 over 2 to the third power. 2-3 = 1/2 3 -n Definition of x : if n is any real number and x is any real number that is not zero, -n x = 1/x n This definition tells us that when we write an exponential expression in reciprocal form, the sign of the exponent must be changed. If the exponent is negative, it is positive in reciprocal form; and if it is positive, it is negative in reciprocal form. We also say x cannot be zero since division by zero is undefined. Example: Simplify a) b) c) d) e) 1/3 -2 -3 3 -2 -3 (-3)-2 -3 -(-3) Answer: a) 9 b) 1/27 c) -1/9 d) 1/9 e) 1/27 We know that x = x x 2 3 We also know that x = x x x Using these definitions we can find the value of the expression 2 3 5 xx=xxxxx=x Product Theorem for Exponents: If m and n and x are real numbers and x ≠ 0 m n xx=x m+n Example: Simplify 2 -5 -4 5 0 x yx y x x Answer: 2 -3 xy Example: Simplify -3 4 5 -10 yy x y x -6 -3 10 2 yx yx Answer: -1 -5 yx We can use the product theorem 2 3 2 2 2 6 to expand (x ) = x x x = x Power Theorem for Exponents: If m and n and x are real numbers m n mn (x ) = x Extension of the Power Theorem: if m a b c are n real mn numbers an bn cn the variables (x y z k …) = x y z k … Example: Simplify -3 2 -2 -3 x(x ) y(xy ) 2 3 -3 2 3 (y ) y (x ) Answer: y x 4 14 If we know the area of a circle, we can find the diameter of the circle and can find the radius of the circle. If we know the circumference of a circle, we can also find the diameter and the radius of the circle. Example: The area of a circle is 12.2 m. What is the approximate circumference of the circle? 2 Answer: πr = Area = 12.2 2 r = 12.2/π 2 r = √(12.2/π) r = 1.97 Circumference = 2πr ≈ 2π(1.97) ≈ 12.38 meters Example: The circumference of a circle is 8π cm. What is the area of the circle? Answer: 8π = 2πr 8π/2π = r 4 cm = r Area = π(4cm) = 16πcm 2 2 HW: Lesson 2 #1-30