Solutions for Session 3, Part B

See solutions for Problems: B1 | B2 | B3 | B4 | B5 | B6 | B7 | B8 | B9

 Problem B1 One way to do this division is to use the rule that xa xb = xa - b. Here, this means that x3 divided by x3 equals x0. The other way to do this is to recognize that we are dividing a number by itself and that any non-zero number divided by itself equals 1: Since the answers must be the same, this means that x0 = 1 for any value of x except 0.

 Problem B2 One way to do this division is to use the rule that xa xb = xa - b. Here, this means that x3 divided by x4 equals x-1. The other way to do this is to write out the numerator and denominator: If x is not 0, we can cancel x three times from the numerator and denominator to leave 1/x as the final answer. Since the answers must be the same, this means that x-1 = 1/x for any value of x except 0 (1/x is undefined for x = 0). Therefore, in any division problem involving a negative exponent, we must restrict the base to a non-zero number.

Problem B3

 a. If x1/2 follows the same rules as xm for integer m, then x1/2 multiplied by x1/2 must be x. This means that x1/2 is the number we multiply by itself to make x. This is the definition of a square root, so x1/2 represents the square root of x. b. Similarly, x1/3 needs to be multiplied by itself three times to make x, so it is the cube root of x. c. If x is a positive number greater than 1, then x1/2 will be greater than x1/3. One way to think about this is to look at x1/6, the sixth root of x. Since x is greater than 1, x1/6 is also greater than 1. If x1/6 were positive but less than 1, multiplying it by itself would give a smaller number. Since x1/6 is greater than 1, multiplying it by itself produces a larger number each time. If you multiply x1/6 by itself, or x1/6, you get x2/6, or x1/3. Meanwhile, x1/6 multiplied by itself three times produces x3/6, or x1/2. Therefore, x1/2 must be larger than x1/3 if x is a positive number greater than 1: Therefore, we know that x0 < x1/6 < x1/3 < x1/2 < x2/3 < x1. (The points on the line above are not drawn to scale.)

 Problem B4 To do this, we write out exponentiation as repeated multiplication: (x3)2 = x3 • x3 = (x • x • x) • (x • x • x). According to the rules given earlier, we add these exponents when multiplying, so the result is x6, the same value as x3 • 2. In general, consider(xa)b. Writing this as a multiplication problem, we'd see xa • xa • xa ... • xa, where there are b occurrences of xa. Adding these exponents, we get a + a + a ... + a; that is, a added to itself a total of b times. Repeated addition is multiplication, so the result is ab, the product of a and b. So we see that, in general, (xa)b = x(a • b) = xab.

 Problem B5 Zero cannot be used as a base for several reasons. The base can only be 0 when working with rules involving multiplication and exponentiation of positive exponents. However, all positive powers of 0 equal 0, and products and sums of 0 are all 0, thus making a one-value system. Since we cannot divide by 0, we cannot define 00 as 0n0n for some n (see Problem B1 for more information). Additionally, we cannot define 0n for any negative n (see Problem B2).

Problem B6

 a. 43,007 = 4.3007 • 104. The lead digit (4) was originally in the 10,000, or 104 position, so when we move it to the 100 position, we must multiply by 104. Multiplying 4.3007 by 104 gives us 43,007. b. 0.00245 = 2.45 • 10-3. The lead digit (2) was originally in the 1/1000, or 10-3 position, so when we move it to the 100 position, we must multiply by 10-3. Multiplying 2.45 by 10-3 gives us 0.00245. c. -675 = -6.75 • 102. The lead digit (6) was originally in the 100, or 102 position, so when we move it to the 100 position, we must multiply by 102. Multiplying -6.75 by 102 gives us -6.75.

Problem B7

 a. 2,300,000 + 790,000 = (2.30 • 106) + (.79 • 106) = 3.09 • 106 b. 10,000,000 • 678,000,000,000 = (1 • 107) • (6.78 • 1011) = 6.78 • 107+ 11 = 6.78 • 1018 c. 1,490,000,000 7,000 = (1.49 • 109) (7 • 103) = (1.49 7) • 109 - 3 = 0.213 • 106

Problem B8

 a. In exponential form, log 100 is 10x = 100. Since 102 = 100, log 100 = 2. b. In exponential form, log3 81 is 3x = 81. Since 3 • 3 • 3 • 3 = 34 = 81, the solution is 4. c. If log x = 4, the equation in exponential form is 104 = x, so x = 10,000. d. The equation is bx = b. This is solved by x = 1 for all valid bases for b (with the convention that b must be positive and not equal to 1).

Problem B9

 a. Log5 50 = x. We can also write this expression as 5x = 50, i.e., 5 raised to what power equals 50? We know that 52 = 25 and 53 = 125, so we can estimate x as somewhere between 2 and 3. To bring our estimate even closer, we can see what happens for x = 2.5, which, converting it into a fraction, can be written as 25/10, or 5/2. So we write 55/2 = = 55.9. Thus, we can further narrow down our estimate by saying that x is slightly under 2.5. b. Similarly, log3 100 = y can be written as 3y = 100. We know that 34 = 81 and 35 = 243, so we can estimate y as somewhere between 4 and 5. For y = 4.5 we can convert 4.5 into a fraction by writing 4.5 = 45/10 = 9/2, we get 39/2 = = 140.3. Thus, we can further narrow down our estimate by saying that y is somewhere between 4 and 4.5.