LESSON 2 Negative exponents Product and power theorems for exponents Circle relationships

Transcription

1 9.A negative eponents LESSON Negative eponents Product and power theorems for eponents Circle relationships.a negative eponents Negative eponents cannot be understood because they are the result of a definition, and thus there is nothing to understand. We define to the third power as follows: = We have agreed that means times times. In a similar fashion, we define to the negative third power to mean over to the third power. = Thus, we have two ways to write the same thing. We give the formal definition of negative eponents as follows: DEFINITION OF n If n is any real number and is any real number that is not zero, n '=' n eample. (a) This definition tells us that when we write an eponential epression in reciprocal form, the sign of the eponent must be changed. If the eponent is negative, it is positive in reciprocal form; and if it is positive, it is negative in reciprocal form. In the definition we say that cannot be zero because division by zero is undefined. (a) (c) (b) (c) (d) ( ) (e) ( ) = = 9 (b) = = 7 Negative signs and negative eponents in the same epression can lead to confusion. If the negative sign is not protected by parentheses, a good ploy is to cover the negative sign with a finger. Then simplify the resulting epression and remove the finger as the last step. problem covered minus sign equivalent epression 9 9 removed finger (d) When we try to slide our finger over the minus sign in (d), we find that we cannot because the minus sign is protected by the parentheses. ( ) problem

2 0 Lesson ( ) protected ( ) 9 equivalent epression (e) One of the minus signs is unprotected. ( ) problem ( ) covered minus sign ( ) equivalent epression 7 7 = removed finger 7.B product We remember that means times theorem = for eponents and means times times = Using these definitions, we can find an epression whose value equals to times. means times which equals 5 This demonstrates the product theorem for eponents, which we state formally in the following bo. PRODUCT THEOREM FOR EXPONENTS If m and n and are real numbers and ' '0, m ' ' n '=' m+n This theorem holds for all real number eponents. eample. y 5 y 5 0 We simplify by adding the eponents of like bases and get y eample. yy y 5 0 y 6 y 0 First we simplify the numerator and the denominator. Then we decide to write the answer with all factors in the numerator. y 6 y = y 5

3 .D circle relationships.c power theorem We can use the product theorem to epand ( ) as for eponents ( ) = = 6 This procedure generalizes to the power theorem for eponents. POWER THEOREM FOR EXPONENTS If m and n and are real numbers, ( m ) n = mn This theorem can be etended to any number of eponential factors. EXTENSION OF THE POWER THEOREM If the variables are real numbers, ( m y a z b k c...) n = mn y an z bn k cn... eample. ( ) y( y ) ( y ) y ( ) First we will use the power theorem in both the numerator and the denominator and get 6 y y y y Now we simplify both the numerator and the denominator, and as the last step, we decide to write all eponential epressions with positive eponents. 8 7 y y 6 = y.d circle relationships eample.5 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. The area of a circle is. m. What is the approimate circumference of the circle? First we find the radius. πr = area πr =. r =. π. r = π r.97 m equation substituted divided by π square root of both sides

4 Lesson We used a calculator and rounded the answer to two decimal places, so the answer is not eact. We indicate that the answer is not eact by using the symbol for approimately equal to. The circumference equals πr, so now we can find the circumference. Circumference = πr equation π(.97) substituted.8 m eample.6 practice The circumference of a circle is 8π cm. What is the area of the circle? First we find the radius. Circumference = πr 8π = πr equation substituted 8π = r divided by π π cm = r Now we can use cm for r to find the area. Area = π r equation = π ( cm) substituted = 6π cm a. b. ( ) c. ( y ) 0 ( y) y 8 y d. The area of a circle is 9π cm. What is the circumference of the circle? problem set. Find.. Find and y. y 8 6. The base of a cylinder is a right triangle topped by a 60 sector of a circle, as shown. If the dimensions are in meters and the height of the cylinder is 8 meters, what is the volume of the cylinder? Find A, B, and C. 5. Find A, B, and C. 80 ( A ) ( B ) ( C ) A ( B ) ( C ) 60

5 problem set 6. The area of the square is 6 cm. What is the length of one side? The circles inside the square are all the same size. What is a radius of one circle? What is the area of one circle? 7. The volume of this circular cylinder 8. The figure shown is the base of a c o n e is 50π cm. What is the height of whose altitude is meters. What is the cylinder? Dimensions are in centi- the volume of the cone? Dimensions meters. are in meters. 5 H Simplify. Write answers with all eponential epressions in the numerator. 9.. ( 0 y ) 5 (y ) 5 0. ( y) 0 y (y ). Simplify. Write answers with positive eponents.. 6. (m ) m ( 0 y ) y. ( y 5 ) ( ) 0 y y 7. m p 0 (m p) m p (m p ) (a b 0 ) ab a b (ab ) (c d) c 5 (c d 0 ) d 5. (b c ) c (b c 0 b ) Simplify. Write answers with negative eponents. 8.. (abc) c b a bc a ( yz ) y 0 (y 0 z ) y 9.. kl k (k 0 L) L k y y ( y) y.. 0. (m n 5 ) m(n 0 ) (m n ) m s ym (s 0 t ) m st 5. [ 0 ( ) ] 6. {[ ( )][ ( )]} 7. { 0 [( 5 )( ) ]} 8. [ 0 7( ) ] 9. Å Å ( 5) 0. Å Å 0 ( )