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Practice Exercises
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This is an open-book quiz You may (and should) refer to the text as you solve these problems Don t hurry! You ll find worked-out answers in App C The solutions in the appendix may not represent the only way a problem can be figured out If you think you can solve a particular problem in a quicker or better way than you see there, by all means try it!
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Practice Exercises
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1 Let x = 23713018568 and y = 0902780337 Find xy to three decimal places using common logarithms 2 Approximate the product xy from Prob 1 using natural logarithms Show that the result is the same as that obtained with common logs when rounded off to three decimal places 3 The power gain of an electronic circuit, in units called decibels (abbreviated dB), can be calculated according to this formula: G = 10 log (Pout/Pin) where G is the gain, Pout is the output signal power, and Pin is the input signal power, both specified in watts Suppose the audio input to the left channel of high-fidelity amplifier is 0535 watts, and the output is 237 watts What is the power gain of this circuit in decibels Round off the answer to the nearest tenth of a decibel 4 Suppose the audio output signal in the scenario of Prob 3 is run through a long length of speaker wire, so instead of the 237 watts that appears at the left-channel amplifier output, the speaker only gets 193 watts What is the power gain of the length of speaker wire, in decibels Round off the answer to three decimal places 5 If a positive real number increases by a factor of exactly 10, how does its common (base-10) logarithm change 6 Show that the solution to Prob 5 is valid for all positive real numbers 7 If a positive real number decreases by a factor of exactly 100 (becomes 1/100 as great), how does its common logarithm change 8 Show that the solution to Prob 7 is valid for all positive real numbers 9 If a positive real number is divided by a factor of 357, how does its natural (base-e) logarithm change Express the answer to two decimal places 10 Show that the solution to Prob 9 is valid for all positive real numbers
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CHAPTER
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Part Three
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This is not a test! It s a review of important general concepts you learned in the previous nine chapters Read it though slowly and let it sink in If you re confused about anything here, or about anything in the section you ve just finished, go back and study that material some more
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21
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Question 21-1
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There are two distinct numbers that, when squared, produce 1 What are those two numbers
Answer 21-1
One of them is the unit imaginary number, which we call j That s the engineer s and applied mathematician s notation Many mathematics texts use the letter i to represent it The other is the negative of the unit imaginary number, j
Question 21-2
How can we show that squaring j produces the same result as squaring j
Answer 21-2
By definition, we know that j 2 = 1 We can multiply the left side of this equation by ( 1)2 without having any effect on its value, because ( 1)2 = 1 When we do that, we get ( 1)2j 2 = 1
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Part Three 499
The power of product rule allows us to rewrite the left side of this equation, so it becomes [( 1)j ]2 = 1 But ( 1)j is the same thing as j, so we can simplify further to get ( j)2 = 1 This is the same result as we obtain by squaring j
Question 21-3
How is an imaginary number put together
Answer 21-3
An imaginary number is the product of j and a real number Suppose we call the real number b If b is positive, then we write the product of j and b as jb If b is negative, we write the product as jb (We always put the minus sign first) If b = 0, then we can write the product as j0 But because j0 = 0, we would more likely write j0 simply as 0
Question 21-4
How can we show that j3 + j7 is the same as j7 + j3, assuming that the familiar distributive law of arithmetic works with the unit imaginary number
Answer 21-4
When we apply the distributive law for multiplication over addition backward to the first expression, we get j3 + j7 = j(3 + 7) The commutative law for addition tells us that 3 + 7 = 7 + 3 By substitution on the right side, we get j3 + j7 = j(7 + 3) Applying the distributive law forward on the right side, we get j3 + j7 = j7 + j3
Question 21-5
How can we show, again assuming that the distributive law works with the unit imaginary number, that if a and b are any two real numbers, then ja + jb is the same as jb + ja
Answer 21-5
We can use the same proof procedure as we did in Answer 21-4, using letter constants instead of numbers! The distributive law tells us that ja + jb = j(a + b)
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