Logarithms FUTURE REVISIONS OF THIS LICENSE

18 Chapter 2 Practice 129 Problem Shew that 100 divides 11 10 − 1 . 130 Problem Shew that 27195 8 − 10887 8 + 10152 8 is divisible by 26460. 131 Problem Shew that 7 divides 2222 5555 + 5555 2222 . 132 Problem Shew that if k is an odd positive integer 1 k + 2 k + ··· + n k is divisible by 1 + 2 + ··· + n. 133 Problem Shew that x + y 5 − x 5 − y 5 = 5xyx + yx 2 + xy + y 2 . 134 Problem Shew that x + a 7 − x 7 − a 7 = 7xax + ax 2 + xa + a 2 2 . 135 Problem Shew that A = x 9999 + x 8888 + x 7777 + ··· + x 1111 + 1 is divisible by B = x 9 + x 8 + x 7 + ··· + x 2 + x + 1. 136 Problem Shew that for any natural number n, there is another natural number x such that each term of the sequence x + 1 , x x + 1 , x xx + 1 , . . . is divisible by n. 137 Problem Shew that 1492 n − 1770 n − 1863 n + 2141 n is divisible by 1946 for all positive integers n. 138 Problem Decompose 1 + x + x 2 + x 3 + ··· + x 624 into factors. 139 Problem Shew that if 2 n − 1 is prime, then n must be prime. Primes of this form are called Mersenne primes. 140 Problem Shew that if 2 n + 1 is a prime, then n must be a power of 2. Primes of this form are called Fermat primes. 141 Problem Let n be a positive integer and x y. Prove that x n − y n x − y ny n−1 . By choosing suitable values of x and y, further prove than € 1 + 1 n Š n € 1 + 1 n + 1 Š n+1 and € 1 + 1 n Š n+1 € 1 + 1 n + 1 Š n+2

2.5 Logarithms

142 Definition Let a 0, a 6= 1 be a real number. A number x is called the logarithm of a number N to the base a if a x = N. In this case we write x = log a N . We enumerate some useful properties of logarithms. We assume that a 0, a 6= 1,M 0,N 0. a log a N = N 2.23 log a MN = log a M + log a N 2.24 log a M N = log a M − log a N 2.25 log a N α = α log a N , α any real number 2.26 log a β N = 1 β log a N , β 6= 0 a real number 2.27 log a blog b a = 1 , b 0, b 6= 1. 2.28 143 Example Given that log 8 √ 2 1024 is a rational number, find it. Solution: We have log 8 √ 2 1024 = log 2 7 2 1024 = 2 7 log 2 2 10 = 20 7 Logarithms 19 144 Example Given that log 2 3 · log 3 4 · log 4 5 ··· log 511 512 is an integer, find it. Solution: Choose a 0, a 6= 1. Then log 2 3 · log 3 4 · log 4 5 ··· log 511 512 = log a 3 log a 2 · log a 4 log a 3 · log a 5 log a 4 ··· log a 512 log a 511 = log a 512 log a 2 . But log a 512 log a 2 = log 2 512 = log 2 2 9 = 9 , so the integer sought is 9. 145 Example Simplify S = log tan 1 ◦ + log tan 2 ◦ + logtan 3 ◦ + ··· + logtan89 ◦ . Solution: Observe that 90 − k ◦ + k ◦ = 90 ◦ . Thus adding the kth term to the 90 − kth term, we obtain S = logtan 1 ◦ tan 89 ◦ + logtan 2 ◦ tan 88 ◦ + logtan 3 ◦ tan 87 ◦ + ··· + logtan44 ◦ tan 46 ◦ + logtan 45 ◦ . As tan k ◦ = 1 tan90 − k ◦ , we get S = log 1 + log 1 + ··· + log1 + logtan45 ◦ . Finally, as tan 45 ◦ = 1 , we gather that S = log 1 + log 1 + ··· + log1 = 0. 146 Example Which is greater log 5 7 or log 8 3? Solution: Clearly log 5 7 1 log 8 3. 147 Example Solve the system 5 € log x y + log y x Š = 26 xy = 64 Solution: Clearly we need x 0, y 0, x 6= 1,y 6= 1. The first equation may be written as 5  log x y + 1 log x y ‹ = 26 which is the same as log x y − 5log y x − 1 5 = 0. Thus the system splits into the two equivalent systems I log x y = 5 , xy = 64 and II log x y = 1 5, xy = 64. Using the conditions x 0, y 0, x 6= 1,y 6= 1 we obtain the two sets of solutions x = 2,y = 32 or x = 32 , y = 2. 148 Example Let TxU be the unique integer satisfying x − 1 TxU ≤ x. For example T2.9U = 2,T− π U = −4. Find Tlog 2 1 U + Tlog 2 2 U + Tlog 2 3 U + ··· + Tlog 2 1000 U. Solution: First observe that 2 9 = 512 1000 1024 = 2 10 . We decompose the interval [1; 1000] into dyadic blocks [1; 1000] = [1; 2[ [ [2; 2 2 [ [ [2 2 ; 2 3 [ [ ··· [ [2 8 ; 2 9 [ [ [2 9 ; 1000] . 20 Chapter 2 If x ∈ [2 k , 2 k+1 [ then Tlog 2 x U = k. If a, b are integers, the interval [a; b[ contains b − a integers. Thus Tlog 2 1 U + Tlog 2 2 U + Tlog 2 3 U + ··· + Tlog 2 1000 U = 2 1 − 2 0 + 2 2 − 2 1 1 +2 3 − 2 2 2 + ··· +2 9 − 2 8 8 +1000 − 2 9 9 = 0 + 2 · 1 + 4 · 2 + 8 · 3 +16 · 4 + 32 · 5+ +64 · 6 + 128 · 7 +256 · 8 + 489 · 9 = 7987 the last interval has 1000 − 512 + 1 = 489 integers. Practice 149 Problem Find the exact value of 1 log 2 1996 + 1 log 3 1996 + 1 log 4 1996 + ··· + 1 log 1996 1996 . 150 Problem Shew that log 1 2 x log 1 3 x only when 0 x 1. 151 Problem Prove that log 3 π + log π 3 2. 152 Problem Let a 1. Shew that 1 log a x 1 only when 1 x a. 153 Problem Let A = log 6 16 , B = log 12 27. Find integers a , b, c such that A + aB + b = c. 154 Problem Solve the equation log 1 3 cos x + √ 5 6 + log 1 3 cos x − √ 5 6 = 2 . 155 Problem Solve log 2 x + log 4 y + log 4 z = 2 , log 3 x + log 9 y + log 9 z = 2 , log 4 x + log 16 y + log 16 z = 2 . 156 Problem Solve the equation x .5 log√ x x2 −x = 3 log9 4 . 157 Problem Given that log ab a = 4, find log ab 3 √ a √ b .

2.6 Complex Numbers