(2/3) 3 ((1 7/8) 2 + 1/2) = (2/3) 3 ((8/8 7/8) 2 + 1/2) (Work from inner parentheses outward) = (2/3) 3 ((1/8) 2 + 1/2) = (8/27) (1/64 + 1/2)

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1 Exponents Problem: Show that 5. Solution: Remember, using our rules of exponents, 5 5, 5. Problems to Do: 1. Simplify each to a single fraction or number: (a) ( 1 ) 5 ( ) 5. And, since (b) / (Simplify 9 / first) (Find common denominator to add fractions) (Divide by multiplying by the reciprocal) (/) ((1 /8) + 1/) (/) ((8/8 /8) + 1/) (Work from inner parentheses outward) (/) ((1/8) + 1/) (8/) (1/6 + 1/) ( a ) n a n (Remember: b b ) n (8/) (1/6 + /6) (Find a common denominator) (8/) (/6) 8 6 (Divide by multiplying by the reciprocal)

2 (c) ( 10 8 ) ( (10 8 ) ) 8 10 (d) ( 10 ) ( 10 ) ( (10 ) ) ( (10 ) ) ( ) (16 10 ) Simplify each expression below as much as possible; your answer should have positive exponents only. (a) (b) (c) (a b ) (a b )(a b 1 ) 1 (a 9 b 6 )(a b )(a b) a 9++ b 6 +1 a 5 b 5 b5 a 5 ( 1 (x 1 y 1 ) 1 (x 1 y 1 ) 1 x + 1 ) 1 (xy) y xy ( 1 + 1) x y xy ( y + x ) xy xy xy y+x xy (xy)(xy) y + x x y y + x (x + y) (x + y) (x y ) (x + y) + (x y 6 ) x + y x y 6

3 (d) ( a ) ( a ) ( a ) ( ) a 6 bc b 1 c b c 8 b c 6 a bc 1 1 a bc 1. Please simplify each of the following expressions as much as possible. (a) 15 / (15 1 ) ( 15 ) 5 5 (b) ( ) (c) 5/ ( 1 ) 5 ( ) 5 5 (d) 81 / ( ) / ( 16 ) 1/ (e) ( 1) 51 Since 51 is odd, ( 1) (f) 15 /5 ( 5 15 ) 5 5. Given that 1 meter 9 inches, what is the distance to the Sun in kilometers if the distance in miles is assumed to be 9 million? Please show all steps leading to your answer and pay close attention to units. We must do several conversions here. To, begin, let us write 9 million as Then, we know that there are 580 feet in a mile. So, the distance to the sun is feet miles feet. Next, we know 1 mile there are 1 inches in a foot. So, the distance to the sun in inches is We are given that 9 inches 1 meter, so inches 1 meter meters. Finally, there are 1000, or 10 meters 9 inches 9 in a kilometer, so, meters 1 kilometer kilometers 9 10 meters kilometers kilometers kilometers Is it true in general that (a) (xy) n x n y n?(n a positive integer) YES (b) (x + y) n x n + y n? NO. Let x 1, y, and n. Then (x + y) n (1 + ) 16, but x n + y n (c) 1/x + 1/y 1/(x + y)? NO. Let x 1 and y. 1/x + 1/y 1 + 1/ /, but 1/(x + y) 1/(1 + ) 1/.

4 Expansion 1. Please expand (x + ) (x + 10) with the result of x + 1x + 0. (x+)(x+10) x x+x 10+ x+ 10 x +10x+x+0 x +1x+0. Please expand (x ) (x 1) and show the result is x 8x +. (x ) (x 1) x x+x ( 1) x ( 1) x x x+ x 8x+. Please expand (x 9) (x + 9) and show the result is x 81. (x 9) (x + 9) x + 9 x 9 x 9 9 x + 9x 9x 81 x 81. What is the expansion of (x 8) (x + 1)? (x 8) (x + 1) x x + 1x 8x 8 1 x + x What is the expansion of (x + 6) (x 9)? (x + 6) (x 9) x x 9x + 6x + 6 ( 9) x x 6. What is the expansion of (x + )? (x + ) (x + )(x + ) x x + x + x + x + 8x What is the expansion of (x a) (x + a)? (x a) (x + a) x x + ax ax a a x a. Note that this is consistent with the result in problem above. 1. In each case expand and collect like terms. (a) x 6 (x + 1) x 6 (x + 1)(x + 1) x 6 (x + (1 + 1)x + 1 1) x 6 (x + x + 1) x 10 + x 8 + x 6 (b) x(x 9)(x + 1) x(x + ( 9 + 1)x + ( 9)(1)) x(x 8x 9) x 8x 9x (c) x (x )(x 1) x (x x x + 1) x x 6 x 5 + x (d) (x + 1) (x + 1)(x + 1)(x + 1) (x + (1 + 1)x + 1 1)(x + 1) (x + x + 1)(x + 1) x + x + x + x + x + 1 x + x + x + 1 (e) x (x 9)(x + 9) x (x 81) x 5 81x 1. Please expand and simplify. We will use Pascal s triangle as before to figure out the coefficients. (a) (x + y) Row of Pascal s triangle has the entries So, (x+y) x + x 6 y + 1x 5 y + 5x y + 5x y + 1x y 5 + xy 6 + y. (b) (x y) 5 Row 5 of Pascal s triangle is: So, (x y) 5 x 5 + 5x ( y) + 10x ( y) + 10x ( y) + 5x( y) + ( y) 5 x 5 15x y + 10x (9y ) + 10x ( y ) + 5x(81y ) y 5 x 5 15x y + 90x y 0x y + 05xy y 5

5 (c) (y + x) Row of Pascal s triangle is: So, (d) (y + x) y + y (x) 1 + 6y (x) + y(x) + (x) y + y (x) + 6y (x ) + y(8x ) + (16x ) y + 8xy + x y + x y + 16x ( x y) ( 1 (x + y)) ( 1) (x + y) (x + y) x + x y + 6x y + xy + y (e) (a + 5b) 5 a 5 + 5a (5b) + 10a (5b) + 10a (5b) + 5a(5b) + (5b) 5 a 5 + 5a b + 10a (5b ) + 10a (15b ) + 5a(65b ) + 15b 5 a 5 + 5a b + 50a b + 150a b + 15ab + 15b 5. What is the coefficient of x 1 y 5 in the expansion of (x + y) 0? (think!) Recall that in the expansion of (x + y) 0, the exponents of each term x m y n should sum to 0. The exponents of the term x 1 y 5 only sum to , so this is not a term in the expansion. Thus, its coefficient is zero.. Please simplify each of the following as much as possible. (a) (b) [(x + h) x ]/h [(x + x h + xh + h ) x ]/h (x h + xh + h )/h [h(x + xh + h )]/h x + xh + h ((x + a) x )/a [(x + x a + 6x a + xa + a ) x ]/a (x a + 6x a + xa + a )/a [a(x + 6x a + xa + a )]/a x + 6x a + xa + a 5

6 Factoring 1. Factor each quadratic below (there are no duds among them). (a) x + x 56 Since we know (x + a)(x + b) x + (a + b)x + ab, we want to find two numbers a and b such that a + b 1 and ab 56. So, we see that we want a 8 and b. Then x + x 56 (x + 8)(x ). Of course, we can always check that we are correct by expanding our answer to see if it matches what we began with. (b) x 81 You should recognize this as the difference of squares: x 81 x 9. So, it factors as (x 9)(x + 9), as we can check. (c) x 6x 91 Remember that we want a and b so that ab 91, so consider the factors of 91: Since we also want a + b 6, we see that we can choose a and b 1. So, the factorization is (x + )(x 1). (d) x + x We always want to factor out any common powers of x first, so we have x + x x(x + ). This is all we can do here. (e) x 6x 0x Again, we first factor out common powers of x: x 6x 0x x(x 6x 0). Then we want to factor the remaining quadratic term. First consider the prime factorization of 0: 0 ***5, so there are several possible choices for a and b here. We could have {a ±, b ±0}, {a ±, b ±10}, {a ±8, b ±5}, etc. Since we also need a + b 6, this leads us to choosing a and b 10, to give us the factorization x 6x 0x x(x + )(x 10).. Factor each below as much as possible. Your answers may be checked by expanding them and comparing to the original expression. (a) x 5x (x 8)(x + ) (b) x 6 100x x (x 100) x (x 10 ) x (x 10)(x + 10) (c) x + x + 10 (x + 5)(x + ) (d) x 11 x 11 (x 11)(x + 11) (e) x 1x + 5 (x 5)(x ) (f) x + 16 Cannot be factored. (g) x + 100x x(x + 100) (h) 5x 15x 5x(x 5) 5x(x 5)(x + 5) (i) x 65 (x ) (5) (x 5)(x + 5) (x 5)(x + 5)(x + 5) (j) 100 x (5 x ) (5 x ) (5 x)(5 + x) 6

7 (k) x + 5x x (x + 5)

8 5 Logs and Exponentials 1. In what follows, suppose that it is given that the following are true: log()., log().8, log(5)., log().85. Use these facts, if applicable, and properties of logs and algebra to provide values for: (no calculators!) (a) log(1000) log(10 ) (b) log(6) log( ) log() + log() (c) log(9) log( ) log() (.8).96 (d) log(1/) log( 1 ) 1 log(). (e) log(/5) log() log(5) (f) log(0) log( 10) log() + log(10) (g) log(81) log( ) log() (.8) 1.9 (h) 10 log() (i) 10 log(5) 10 log(5 1) (j) 100 log() (10 ) log() 10 log() 10 log() 9 (k) 1000 log() (10 ) log() 10 log() 10 log() 8 8

9 6 Series 1. Write out each series below. (a) (b) (c) (d) (e) i i1 5 i i k k (k + ) ( + ) + ( + ) + ( 5 + ) + ( 6 + ) + ( + ) k 1 i Please sum the following expressions: (a) First, note that each term in this expression is divisible by, so we can rewrite this as ( ). Now, we can see that this 100 is just the arithmetic series i1 i, so the sum is ( ) (b) 1 + 1/ + 1/9 + 1/ + + 1/18 ( 1 )0 + ( 1 )1 + ( 1 ) + ( 1 ) + + ( ) i 1 ( 1 ). This is a geometric series, so its sum is 1 ( 1 ) (c) i j j5 80 j j j j

10 (d) ( ) 1 ( ) ( [ (1 1 ) ( ) i i0 ( ) ( 1 5 ) ) ( ) ( ) 5 ( ( ) ( ) ( ) ) 6 ( ) ] (e) + / + / + /8 + /16 + / + /6 + /18 + /56 + /51 + /10 [ (1 ) 0 ( ) 1 ( ) ] ( ) i 1 i0 ( ( 1 1 ) 11 ) Please sum the following series: (a) 6 ( ) i / + 1/ /9 i0 (1 ( 1 ) ) (b) 10 ( 1 1/ + 1/ 1/8 + 1/ /10 1 ) i i0 (1 ( ) 11)

11 (c) 5 + 5/ + 5/ + + 5/ i0 ( ) i 1 (1 ( 1 ) 9) (d) 1/5 + 1/15 + 1/ / i ( 1 5 ) (e) ( ) 50 k k [ 50 ] k (1 + ) 5 k Consider the following series: (i (i + 1) ) i1 (a) Write out the first three terms of the series and also the last two. Terms: i 1 : i : i : i 99 : i 100 : (b) See if you can sum the series. Sometimes series can be summed easily by grouping the terms in a careful way. (1 )+( )+( )+ + ( )+( ) 1 +( + )+( + )+ +( ) 101. Now it is easy to see that the sum is just Let s look at the odd positive integers and add them up. For example,

12 (a) Can you find a method for adding up the first 1 odd numbers? (other than simply adding them one by one) We can think of this as Gauss did. Write the first 1 odd numbers forward, and then backward and let s see what we get. So, the sum is since there are 1 pairs of numbers that sum to, counting each number in the series twice. (b) Can you extend your method and find a formula for the sum of the first n odd integers? The sum of the first n odd integers can be n written as: k 1. Using the idea above, we see that the first and last k1 number in this series should sum to 1 + (n 1) n. Then, our idea is that there are n pairs of numbers summing to n. But, this counts each number in our series twice, so the sum is (n)(n) n. So, the sum of the first n odd integers is just n. 5. A very hard ball is dropped from a height of 100 feet. It strikes the ground and bounces back up, each time reaching a height of / the height of the previous bounce. It bounces 1 times before coming to rest. What is the total distance it traveled? What does this have to do with series?? When the ball is first dropped, it travels 100 feet to the ground. After it bounces, it travels up 100 feet, then down 100 feet before bouncing again. When the ball bounces this time, it only reaches a height of ( 100) ( ) 100 feet before falling to the ground again. So, generalizing we see that the height the ball reaches after the n th bounce is ( ) n 100 feet. So the total distance the ball travels is: ( ) ( ) ( ) feet. Now, we can use the techniques we ve learned to find the sum ( ) ( ) ( ) 1 1 ( ) n n ( ) n n ( 1 )