How do you determine log values for 2, 3, 4…?
Johnny: I was happy using the semilog paper but I still have a question about how on a log sclale they come up with the lines for 2, 3, 4 and so on.
Sara: We will have to read up on that. I think we have a chapter on logs in our book. Why don’t we do this another time?
With this Johnny went home happy that Sara’s mom’s magic semi-log paper had helped him. He glanced through his algebra book quickly but then got busy with other things. Next morning, Sara and Johnny walked to school as usual when Johnny brought up the discussion from the previous day. However, Sara suggested that they should talk about it later. After school, they walked back together and Johnny asked if she wanted to come over to his place for a while. Of course, Sara was always happy to be with the boy she loved, and she agreed. They went over to Johnny’s place, had some pop and watched the TV for a while.
Sara looked at her watch and then said: Johnny, I have to be home in an hour because our family is going out for a dinner. So we better finish the discussion from the last night.
Johnny: I looked at the algebra book but things made no sense to me. So I stopped.
Sara: That’s okay but I want to see how much you remember from your middle school years.
Johnny: I was always poor at Math but go ahead.
Inverse functions
Sara: Do you remember what is the opposite of addition?
Johnny: I am not that dumb. Of course, I remember that the opposite of addtion is subtraction. I remember reading about the inverse functions like multiplication and division, and square and square root but what does that have to do with the exponents?
Sara: The inverse of an exponent is log. We can either write
y = 10m or m = log10 y. It is the same thing.
Johnny: Why did you write log10 instead of log?
Sara: Because we use a decimal system for numbers, it is convenient to think of numbers 10n rather than exponents with other bases. So routinely we think of log of a number to the base 10 and write it as log10. So, from now on when I write just log, it will mean that it is really log10.
Johnny: How do exponent rules apply to logs?
Sara: 100 = 102, so log 100 = 2
Johnny: You mean that log 1000 will equal 3. 100 x 1000 = 105 and then log 105 = 5. We know the rule 10m x 10n = 10m+n . Because log 10m = m and log 10n = n, it would mean that log (10m x 10n) = m + n. I don’t know why our book did not explain it in this simple way. This makes lot more sense.
Sara: You wrote the exponent with base 10, we could use any base and write
am x an = am+n. Therefore log (am x an) = log am+n or log (am x an) = log am + log an. The log am can also be writen as mlog a.
Johnny: Could we also write: log (j x k) = log j + log k?
Sara: Yes, we could write this even for longer products:
log (j x k x l,,,,) = log j + log k + log l……,
and here is the bonus log (j/k) = log j – log k.
Johnny: Does that mean, instead of those hard multiplications and divisions of numbers, we can take the logs and sum them up or subtract them.
Sara: Yes, my mom told me that calculators were not allowed in their high school exams but they were allowed to use tables of logarithms for complicated multiplication and divisions.
Johnny: We didn’t cover the exponent rule of (am)n, but to me it makes sense that the log of this exponent will become n log (am).
Sara: Genius. I don’t know why the log chapter in the book didn’t make sense to you. You didn’t read it, did you?
Values of logs of 2,3,4…
Johnny: We have one more thing to figure out. How do they figure out where to draw the lines for the numbers between 1 and 10 in the semilog paper? You said you had some idea.
Sara: I don’t know how they calculated them. They must have sophisticated methods but let’s do this approximation for logs of 2,3,4,5 and 6.
Johnny: How?
Sara: Do you remember that when working with exponents we said
2 x 2 x 2 x 2 x 2 = 32 and 32 x 32 =1024 which we rounded off to 1000.
Johnny; I get it. 210 ≈ 1000. Therefore, log (210) = log (1000) or 10 log 2 ≈ 3 or log 2 =3/10 ≈ 0.3. Wow, that’s how you figure out the first line on the magic graph paper?
Sara: I just checked using my calculator the actual value of log 2 is 0.3010. So you are right, genius.
Johnny: Now, for log 3, I could write log (32) = log 25 or 5 log 2. Therefore,
log (32) = 5 x 0.3010 = 1.505, and then log 3.2 = 1.505 -1 = 0.505. So log 3 will be slightly less than 0.505. What is its actual value?
Sara: log 3 is 0.4771. So you were very close. For 4 it is easy 4 = 22
So log 4 = 2 log 2 = 0.6020. Now, how will you determine log 5?
Johnny: I am not dummy. They taught me in my kindergarten that
5 x 2 =10. So, I will go log 5 + log 2 = log 10 = 1 or log 5 = 1 – 0.3010 or 0.6990.
Sara: log 6 will also be easy because 6 = 3 x 2. So log 6 = log 3 + log 2.
Johnny: I am impressed that we can figure these out. Mathematicians with years of experience must be able to figure them out more accurately.
Sara: So, is that it?
There is no zero on the semilog scale. Why?
Johnny: Something is bothering me. There is no zero on the semilog scale.
Sara: I think I can explain that. Let’s see, log 10 = 1, log 1/10 = -1, log 1/100 = -2 and so on. So log 0 = log (1/10)infinity = minus infinity. So you will need an infinitely long semilog paper to show log (0).
Johnny: All the logs we did were for the base 10. How would they work on a different base and how would you convert them to log10.
Sara: It is easy to do with log2. By definition log2 2 = 1 because 21 =2. We just figured out that log10 = 0.3010. So for any number m, we could write that log10 m = 0.3010 log2 m. You could the same thing in general and say
logn m = log10 m = lognm x log10 n.
Johnny: There was a word called natural logarithm in the book. What is so natural about them?
Sara: When the base is a constant called e, the logarithm is called natural. I don’t know what is natural about e but it is called Euler’s constant and has a value of just over 2.7.
Johnny: Did you ask your mom?
Sara: Yes but she said that it will be better to learn it later when we study binomials in our algebra class.
Johnny: I am sure there’s lots more to learn but I think I have the basics now. Thanks Sara.
Sara went home for the family dinner. It was easy for Johnny and Sara to follow the lessons on log in the class, and do all the homework.
Challenge
Mr. Johnson was proud of his basic knowledge of arithmetic and often taunted his kids for not having these skills until one day his daughter who was in high school now asked him this question. There are 64 squares on a chess board. Approximately, how much money would you need to put in the last square if you were to put one dollar in the first square, two in the next and keep doubling them until all the squares are filled?
Mr. Johnson struggled with the problem for two hours already, can you help Lisa instead?
Solution: Dollars in square 1 = 1 = 20
Dollars in square 2 = 2 = 21
Dollars in square 3 = 2 x 2= 22
Continuing this series dollars in square n = 2(n-1)
For the last square n =64 and the dollars in it will be 263.
For a quick answer log 263 = 63 log 2 = 63 x 0.3010 = 18.963 or rounded of to 19.
So the dollar amount in the last square will be inverse log 19 = 1019 =
10 x 1,000,1000,1000,1000,1000,1000.
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