Homework Problem Set Sample Solutions S.9. Below are formulas Britney Gallivan created when she was doing her paper-folding extra credit assignment. his formula determines the minimum width, WW, of a square piece of paper of thickness needed to fold it in half nn times, alternating horizontal and vertical folds. his formula determines the minimum length, LL, of a long rectangular piece of paper of thickness needed to fold it in half nn times, always folding perpendicular to the long side. WW = ππ 3(nn ) LL = ππππ 6 (nn + 4)( nn ) Use the appropriate formula to verify why it is possible to fold a inch by inch sheet of gold foil in half times. Use 00. millionth of a meter for the thickness of gold foil. Given that the thickness of the gold foil is 0.8 millionth of a meter, we have 0.8 00 cm m,000,000 m = 0.00008 cm in = 0.00000 in..54 cm Using the formula with n = 3 and = 0.00000, we get W = π 3(n ) W = π(0.00000) 3(3 ) 9. hus, any square sheet of gold foil larger than 9. inches by 9. inches can be folded in half 3 times, so a 0 inch by 0 inch sheet of gold foil can be folded in half 3 times. S.0. Use the formula from Problem to determine if you can fold an unrolled roll of toilet paper in half more than times. Assume that the thickness of a sheet of toilet paper is approximately 00. 000000 iiii. and that one roll is ffff. long. First, convert feet to inches. 0 ft. = 4 in. hen, substitute 0.00 and 0 into the formula for and n, respectively. L = π(0.00) ( 0 + 4)( 0 ) = 0.3 6 he roll is just long enough to fold in half 0 times. Unit 3: Exponential Functions 49
3. Suppose your class tried to fold an unrolled roll of toilet paper that was originally 44 iiii. wide and 3333 ffff. long. oilet paper is approximately 00. 000000 iiii. thick. Complete each table, and represent the area and thickness using powers of. hen write an equation for the thickness and the area of the top. Number of Folds, nn 0 hickness After nn Folds (iiii.) 00. 000000 = 00. 000000 00 Number of Folds, nn 0 Area on op After nn Folds (iinn ) = 00 00. 000000 = 00. 000000 777777 = 00. 000000 = 00. 000000 333333 = 3 00. 000000 = 00. 000000 33 3 = 33 4 00. 000000 = 00. 000000 44 4 9999 = 44 5 00. 000000 = 00. 000000 55 5 4444 = 55 6 00. = 00. 000000 66 6. 55 = 66 n 00. 000000 nn, where nn is a nonnegative integer. n nn S. 4. Apply the properties of exponents to rewrite each expression in the form kkxx nn, where nn is an integer and xx 00. a. (xx 33 ) 33xx 55 (66xx) 3 36x 3+5+ = 6x 0 b. 33xx 44 ( 66xx) 3xx 4 36xx = 08xx 6 Unit 3: Exponential Functions 50
c. xx 33 xx 55 33xx 44 3 xx 3+5 4 = 3 xx d. 55(xx 33 ) 33 (xx) 44 5 6 xx 9+( 4) = 5 6 xx 3 e. xx 44xx 33 xx 6 4 3 xx 3 = 64xx 6 3 = 64xx 9 Unit 3: Exponential Functions 5
S. 5. Jonah was trying to rewrite expressions using the properties of exponents and properties of algebra for nonzero values of xx. In each problem, he made a mistake. Explain where he made a mistake in each part, and provide a correct solution. Jonah s Incorrect Work A. (3xx ) 3 = 9xx 6 B. = 6xx5 3xx 5 C. xx xx3 3xx = 3 xx3 MP.3 In Part A, he multiplied 3 by the exponent 3. he correct solution is 3 3 x 6 = 7 x 6. In Part B, he multiplied by 3 when he rewrote x 5. he 3 should remain in the denominator of the expression. he correct solution is 3 x5. In Part C, he only divided the first term by 3x, but he should have divided both terms by 3x. he correct solution is x x3 = x. 3x 3x 3 3 6. Apply the properties of exponents to verify that each statement is an identity. A. nn+ 3 nn = 3 nn for integer values of nn nn+ 33 nn = nn 33 nn nn = 33 nn = nn 33 B. 3 nn+ 3 nn = 3 nn for integer values of nn 33 nn+ 33 nn = 33 nn 33 33 nn = 33 nn (33 ) = 33 nn = 33 nn Unit 3: Exponential Functions 5
C. (3 nn ) 4nn 3 = 3 3 nn for integer values of nn 44nn (33 nn ) 33 = 33 33 nn = 33 33 = 33 33 7. If xx = 55aa 44 and aa = bb 33, express xx in terms of bb. By the substitution property, if x = 5a 4 and a = b 3, then x = 5(b 3 ) 4. Rewriting the right side in an equivalent form gives x = 80b. 8. If aa = bb 33 and bb = cc, express aa in terms of cc. By the substitution property, if a = b 3 and b = c, then a = c 3. Rewriting the right side in an equivalent form gives a = 4 c 6. 9. If xx = 33yy 44 and yy = ss xx 33, show that ss = 5555yy. S.3 Rewrite the equation y = s x 3 to isolate the variable s. y = s x 3 x 3 y = s By the substitution property, if s = x 3 y and x = 3y 4, then s = (3y 4 ) 3 y. Rewriting the right side in an equivalent form gives s = 7y y = 54y 3. 0. Do the following tasks without a calculator. a. Express 88 33 as a power of. 88 33 = 33 33 = 99 b. Divide 44 by. 44 = 3333 = or 44 44 44 = = 44 55 Unit 3: Exponential Functions 53
. Use powers of to perform each calculation without a calculator or other technology. a. 77 55 77 55 = 77 55 44 = 77+55 44 = 88 = b. 555555555555 333333 555555555555 333333 = 555555 3333 = 99 55 = 44 =. Write the first five terms of each of the following recursively defined sequences: a. aa nn+ = aa nn, aa = 33 33, 66,,, 4444 b. aa nn+ = (aa nn ), aa = 33 33, 99, 8888, 66666666, 4444 000000 777777 c. aa nn+ = (aa nn ), aa = xx, where xx is a real number Write each term in the form kkxx nn. xx, xx, 88xx 44, xx 88, 3333333333xx d. aa nn+ = (aa nn ), aa = yy, (yy 00) Write each term in the form kkxx nn. S.4 CHALLENGE yy, yy, yy, yy, yy 3. here is an identity that states ( rr)( + rr + rr + + rr nn ) = rr nn, where rr is a real number and nn is a positive integer. Use this identity to respond to parts (a) (g) below. a. Rewrite the given identity to isolate the sum + rr + rr + + rr nn for rr. ( + rr + rr + + rr nn ) = rrnn rr b. Find an explicit formula for + + + 33 + +. = Unit 3: Exponential Functions 54
c. Find an explicit formula for + aa + aa + aa 33 + + aa in terms of powers of aa. aa aa d. Jerry simplified the sum + aa + aa + aa 33 + aa 44 + aa 55 by writing + aa. What did he do wrong? He assumed that when you add terms with the same base, you also add the exponents. You only add the exponents when you multiply terms with the same base. e. Find an explicit formula for + aa + (aa) + (aa) 33 + + (aa) in terms of powers of aa. () f. Find an explicit formula for 33 + 33(aa) + 33(aa) + 33(aa) 33 + + 33(aa) in terms of powers of aa. Hint: Use part (e). () 33 g. Find an explicit formula for PP + PP( + rr) + PP( + rr) + PP( + rr) 33 + + PP( + rr) nn in terms of powers of ( + rr). ( + rr)nn ( + rr)nn PP = PP ( + rr) rr Unit 3: Exponential Functions 55