



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 x^{a} x^{b} = x^{a  b}. Here, this means that x^{3} divided by x^{3} equals x^{0}. The other way to do this is to recognize that we are dividing a number by itself and that any nonzero number divided by itself equals 1:
Since the answers must be the same, this means that x^{0} = 1 for any value of x except 0.
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Problem B2  
One way to do this division is to use the rule that x^{a} x^{b} = x^{a  b}. Here, this means that x^{3} divided by x^{4} 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 nonzero number.
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Problem B3  
a.  If x^{1/2} follows the same rules as x^{m} for integer m, then x^{1/2} multiplied by x^{1/2} must be x. This means that x^{1/2 }is the number we multiply by itself to make x. This is the definition of a square root, so x^{1/2} represents the square root of x. 
b.  Similarly, x^{1/3} 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 x^{1/2} will be greater than x^{1/3}. One way to think about this is to look at x^{1/6}, the sixth root of x. Since x is greater than 1, x^{1/6} is also greater than 1. If x^{1/6} were positive but less than 1, multiplying it by itself would give a smaller number. Since x^{1/6} is greater than 1, multiplying it by itself produces a larger number each time.
If you multiply x^{1/6} by itself, or x^{1/6}, you get x^{2/6}, or x^{1/3}. Meanwhile, x^{1/6} multiplied by itself three times produces x^{3/6}, or x^{1/2}. Therefore, x^{1/2} must be larger than x^{1/3} if x is a positive number greater than 1:
Therefore, we know that x^{0} < x^{1/6} < x^{1/3} < x^{1/2} < x^{2/3} < x^{1}.
(The points on the line above are not drawn to scale.)

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Problem B4  
To do this, we write out exponentiation as repeated multiplication: (x^{3})^{2} = x^{3} • x^{3} = (x • x • x) • (x • x • x). According to the rules given earlier, we add these exponents when multiplying, so the result is x^{6}, the same value as x^{3 • 2}.
In general, consider(x^{a})^{b}. Writing this as a multiplication problem, we'd see x^{a} • x^{a} • x^{a} ... • x^{a}, where there are b occurrences of x^{a}. 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, (x^{a})^{b} = x^{(a • b)} = x^{ab}.
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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 onevalue system. Since we cannot divide by 0, we cannot define 0^{0} as 0^{n}0^{n} for some n (see Problem B1 for more information). Additionally, we cannot define 0^{n} for any negative n (see Problem B2).
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Problem B6  
a.  43,007 = 4.3007 • 10^{4}. The lead digit (4) was originally in the 10,000, or 10^{4} position, so when we move it to the 10^{0} position, we must multiply by 10^{4}. Multiplying 4.3007 by 10^{4} 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 10^{0} position, we must multiply by 10^{3}. Multiplying 2.45 by 10^{3} gives us 0.00245. 
c.  675 = 6.75 • 10^{2}. The lead digit (6) was originally in the 100, or 10^{2} position, so when we move it to the 10^{0} position, we must multiply by 10^{2}. Multiplying 6.75 by 10^{2} gives us 6.75. 
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Problem B7  
a.  2,300,000 + 790,000 = (2.30 • 10^{6}) + (.79 • 10^{6}) = 3.09 • 10^{6} 
b.  10,000,000 • 678,000,000,000 = (1 • 10^{7}) • (6.78 • 10^{11}) = 6.78 • 10^{7+ 11} = 6.78 • 10^{18} 
c.  1,490,000,000 7,000 = (1.49 • 10^{9}) (7 • 10^{3}) = (1.49 7) • 10^{9  3} = 0.213 • 10^{6} 
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Problem B8  
a.  In exponential form, log 100 is 10^{x} = 100. Since 10^{2} = 100, log 100 = 2. 
b.  In exponential form, log_{3} 81 is 3^{x} = 81. Since 3 • 3 • 3 • 3 = 3^{4} = 81, the solution is 4. 
c.  If log x = 4, the equation in exponential form is 10^{4} = x, so x = 10,000. 
d.  The equation is b^{x} = 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). 
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Problem B9  
a.  Log_{5} 50 = x. We can also write this expression as 5^{x} = 50, i.e., 5 raised to what power equals 50? We know that 5^{2} = 25 and 5^{3} = 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 5^{5/2} = = 55.9. Thus, we can further narrow down our estimate by saying that x is slightly under 2.5. 
b.  Similarly, log_{3} 100 = y can be written as 3^{y} = 100. We know that 3^{4} = 81 and 3^{5} = 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 3^{9/2} = = 140.3. Thus, we can further narrow down our estimate by saying that y is somewhere between 4 and 4.5. 
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