Exponents and Logarithms Discussion

Student: So to find the dimension of a fractal I need to solve:

for D How do I do this?

Mentor: We need to use some mathematical manipulation to either guess at the exponent D or move D out of the exponent. Let's try guessing first. What can you tell me about the size of D for the Koch Snowflake?

Student: Let's see; the number of pieces we replace the original line segment with is 4, and they are each 1/3 of the original segment long, so S = 3 and N = 4.

What power of 3 gives 4? Well 31 = 3 and 32 = 9, so D must be somewhere between 1 and 2.

Mentor: Where between 1 and 2? Let's use our calculators and make a table.

D = 1.5 3D = 5.196152 -- Too big
D = 1.3 3D = 4.171168 -- Too big
D = 1.2 3D = 3.737193 -- Too small
D = 1.25 3D = 3.948222 -- Too small but closer!

Student: That could take a long time! But at least we know D is somewhere near 1.25.

Mentor: There is a more direct way: The answer can be found using a property of logarithms:

What is going on here? We are taking the "log" of both sides of an equation. Rule number 1 for logs: Taking the log of both sides is allowed. Next we are swinging the exponent (b) down in front of the log(a) and multiplying instead of exponentiating. This is rule number 2: Logging an exponent is equal to multiplying the log by the exponent instead.

Logs are used so often that most calculators have a button built in to calculate logs. So we simply calculate log(4) / log(3) to get 1.261860!

See? The Koch Snowflake is not one dimensional or two dimensional but somewhere in the middle!

Try some of these yourself: Fractal Dimensions

Student: So, did someone invent logs just to find fractal dimension?

Mentor: No! Logarithms are very old, and until calculators came along and made arithmetic easy, they were used to make multiplying and dividing large numbers faster. John Napier invented them for this very purpose in the early seventeenth century, publishing a book entitled A Description of the Wonderful Law of Logarithms in 1614. Here is the our modern day definition for a log (base 10):

This says that the log base 10 of y is the power of 10 that is equal to y. Here are some logs we can find in our heads:

log(10) = 1 because 10 1 = 10
log(100) = 2 because 102 = 100
log(1/10) = -1 because 10-1=1/10

Others are harder. What is log(5)? What power of 10 gives 5? It must be some exponent larger than 0 but smaller than 1. How do we find it? Until we learn some calculus (which is a while from now!) we'll rely on our calculator. Log(5) = 0.698970004.

Student: You said that logs were used to multiply and divide large numbers. How did that work?

Mentor: Let's suppose we have two numbers to multiply, say: 1,234,567 and 7,654,321. Let's also suppose that we have a book with a table that we can look up logs in, and we find that

log(1,234,567) = 6.091515 and log(7,654,321)=6.883907.

So to multiply these numbers we can either calculate

1,234,567 x 7,654,321


106.091515 x 106.883907.

Which is easier?

Student: Well, I think that the second one might be better since when I have to multiply two powers together and the bases are the same, I add the exponents! So, since 6.091515+6.883907 = 12.975422, the answer is 1012.975422. What do I do with that?

Mentor: If we had that table of logs, we could look 12.975422 up in the table and give the answer. We have turned multiplying two large numbers into adding two not so large numbers! Of course, this idea might not be as interesting now that we have calculators to do these multiplications for us. In this case we get 1012.975422= 9,449,786,575,000.

Student: Wait! When I do 1,234,567 x 7,654,321 on my calculator, I get 9.449772114E12. Is that the same as 9,449,786,575,000?

Mentor: No, it's not. Let's be very precise here. The logs given above were approximates -- log(1,234,567) is about 6.091515 -- so if we multiply numbers out this way, we are getting an estimate of the answer. Your number was in scientific notation; that E12 on the end says to move the decimal point 12 spaces right: 9,449,772,114,000 is what your calculator gives. That is also only an estimate. Can you explain how I know that it is not exact?

Student: Well, if I multiply out the two numbers by hand, I know that last digit will be a 7, because the last digit of 1,234,567 is 7 and the last digit of 7,654,321 is 1 -- and 1 x 7 = 7.

Mentor: Good! I should mention one more thing about logs before we finish up: Using 10 for the base is not the only choice. In fact, any positive number can be used as the base. The most popular are 10, 2 and e. E is a special number that appears in science -- when we use the log base e we call it the Natural Log (ln). Later in science and math classes, you'll find out that e and natural logs are used a lot.