What the Heck are Logarithms?

WARNING 1: DO NOT use your calculator when studying this handout. Use of the calculator will probably make you miss the point.

WARNING 2: The FIRST RULE for using a calculator is this:
NEVER BELIEVE THE ANSWER THE CALCULATOR GIVES YOU.

Whenever possible, ESTIMATE the answer. If the calculator's result is not reasonably close to your estimate, you should be suspicious; you may have pushed the buttons wrong. Also, it is a good idea to check your work by another method whenever possible. Some examples will be given later. You will also discover ways to make fairly good estimates of logarithms later in this handout, but for now, on with the show.

Many college freshmen have the notion that logarithms are a button on calculators that magically comes up with a value that in some mysterious way allows them to solve certain types of problems. The purpose of this paper is to help you develop an intuitive understanding of what logarithms are and how to use them. If you can estimate within 0.02 (without any device other than your own brain) the value of log 370 then you can probably just scan this paper and go about your business. If you are not confident that you can make a reasonable estimate (or don't even know what the question means) read on: a relatively small amount of thought about the questions and answers given here should lead you toward a comprehension of logarithms that will prove quite valuable throughout your studies and work in engineering.

QUESTIONS Remember: no calculators - brains only! Also, when you finally run across a question you cannot answer, don't just look up the answer (all answers are given at the end of this web page), THINK about it first. Review the previous questions, read the hints, don't give up and look prematurely; you will only be cheating yourself.

Question Group A

1. If 10x = 100, what is x?

I feel sure everyone can answer that one easily.

2. If 10x = 100000, what is x?

Another easy one.

3. If 10x = 4873, estimate x. (Remember - no calculators! You should be able to get within a few tenths at least.)

This one is a bit harder. Can you at least come up with an upper and lower limit for x? (Hint: consider 10x = 1000 and 10x = 10000.)

I imagine most of the students reading this have had little trouble so far. Now let's consider another type of question. (Remember: NO CALCULATORS!)

4. If X = Log10 1000, determine X.

(Note: Log10 means a base 10 logarithm. More about logarithms using other bases shortly.)

Here is the point where many students either hesitate or draw a blank. After all, you have been taught to get logarithms from a calculator, and calculators are off-limits for this problem. This is what is really silly: This question is no harder than question 1, and you could all answer that one.

Here is your hint:
"If X = Log101000, determine X" asks EXACTLY the same question as "If 10x = 1000, what is x?"
Log10 1000 asks, "Find a number such that if I raise 10 (the base of the logarithm) to that number, the result is 1000."

Don't proceed until you are sure you understand question 4, then try another easy one - question 5. You should not have to hesitate much before answering this next one.

NOTE: Henceforth, the subscript 10 will be omitted when base 10 logarithms are used. Thus Log 33 means Log10 33. This notation is the same as that used on your HP calculator.

5. If X = Log 1000000, determine X.

If you still hesitated more than a few seconds, think about the explanation below question 4 some more before proceeding. IMPORTANT: If you still cannot figure it out, get help before proceeding further.

6. If X = Log 5.5, determine X.

Again, can you determine upper and lower limits for X? If you could answer question 3, this one is the same type of question! Later in this tutorial, I will show you a simple way to narrow the range of the upper and lower limits to improve your estimates significantly.

If you are confident that you got the first six problems correct, you might wish to check the answers to question group A at this point just to make sure.

QUICK REVIEW 1: Log Q is a number such that if ten is raised to this number, the result is Q. In other words:

10Log Q = Q or Log 10X = X

(The second equation can be stated as: find a number such that if 10 is raised to that number the result is 10X. The answer is rather obviously X.)

Those two equations are quite important. Be sure you understand them. You will see them again in substantially harder problems.

Next, a few important properties of logarithms. Most of these you were probably forced to memorize at some point in High School.

MEMORIZATION IS DOUBLE-PLUS-UNCLEVER!

Things you memorize tend to be stored in short term memory (although frequent use keeps refreshing short-term memory, rather like dynamic RAM in a computer), whereas concepts which are UNDERSTOOD tend to be stored more permanently. Also, if you UNDERSTAND basic concepts, you don't have to clutter up your brain with memorized rules to deal with all conceivable situations; you can derive the relationships you need from your understanding. This will be demonstrated at several points in this tutorial.

Next, consider the logarithm of a product of two numbers.

As an example, let X = Log (100 . 10). This is obviously the same as Log 1000 = 3. Now consider the individual logarithms of the two numbers being multiplied.

Log 100 = 2 and Log 10 = 1.

Note that at least in this case Log (100 . 10) = Log 100 + Log 10 since 3 = 2 + 1. Let's see if we can determine a general rule about the logarithm of products from our basic understanding of logarithms and powers.

For the moment, we will consider positive integer powers of ten only in order to simplify the thought process. (You may actually have to think a bit to get through this, but persevere; it is not that difficult.)

First, we know that 10A means "multiply A tens together".

Therefore 10A . 10B means "multiply A tens together then multiply that by B more tens". In other words, we have multiplied together a total of (A + B) tens. We can therefore say that 10A . 10B = 10(A+B). From this it follows that Log (10A . 10B) = Log (10(A+B)). Since we know that Log 10X = X, this implies that:

Log (10A . 10B) = Log (10(A+B)) = A+B = Log 10A + Log 10B

Stated in English, the logarithm of a product is the sum of the logarithms of the multiplicands. Although we assumed above that A and B were both positive integers, there is no reason to place this restriction on these values except for ease in thinking about it. Since any positive value can be expressed as a power of ten, (in general yielding a fractional exponent) This rule can be generalized to state

Log (Q . R) = Log Q + Log R

CHALLENGE 1:

The complete derivation can be done as an exercise. Start with Q =10Log Q. If you can derive the formula above (Log (Q . R) = Log Q + Log R), you are well on your way to understanding logarithms.

Another related property is that the logarithm of a number raised to a power equals the power times the logarithm of the number. In other words:

Log XA = A Log X

As examples, Log 1017 = 17 Log 10

Log 845 = 5 Log 84

Log 34.8-= -Log 34.8

CHALLENGE 2: (definitely optional)

Derive this relationship using only the concepts and relations previously stated. This one is significantly harder than Challenge 1. Hint: Begin by considering only positive integer values of A. First derive the special case for A = 1. Next derive the case for A=2. Finally, derive the case for A+1 assuming you have the case for A. If you succeed in deriving this for integer values of A, see if you can extend it to fractional values and negative values.

QUICK REVIEW 2: Recall again that 10Log Q = Q and Log 10X = X.

The two new properties are:

Log (A . B) = Log A + Log B (Log of product equals sum of Logs)

Log XA = A Log X (Log of power equals power times Log)

HANDY HINT: If you can remember that Log 20.3, you can impress people with your ability to estimate base 10 logarithms. For example,

Log 800 = Log (8 . 100) = Log 8 + Log 100 = Log 23 + Log 100 =

3 Log 2 + Log 100(3 . 0.3) + 2 = 2.9

If you can remember (or calculate in your head) the first several powers of 2, (i.e. 2, 4, 8, 16, 32, 64, 128, 256, 512) you can arrive at quite good estimates of logarithms without a calculator. For example: Log 7147

We can establish that

Log (64 . 100) < Log 7147 < Log (8 . 1000) thus

6 Log 2 +2 < Log 7147 < 3 Log 2 + 3 thus

3.8 < Log 7147 < 3.9 thus a reasonable estimate would be

Log 71473.85 (The actual value is 3.85412...)

Question Group B

Question 7: Estimate the following logarithms. (Remember: NO CALCULATORS!)

A. Log 3592

B. Log 305

C. Log 0.0423 (remember that 10-1 = 0.1, 10-2 = 0.01, etc.)

You might also reconsider problems 3 and 6, using this method to improve your estimates. You may see how I estimated these in the answers to Question Group B.

We have now covered the basics of base ten logarithms. The concept of a logarithm is more general than this, however. Logarithms can have any base: except for two special cases, the notation usually uses a subscript after Log and the meaning is slightly modified. For a base r logarithm of X, the notation is Logr X and this means "Find a number such that if r is raised to that number, the result is X."

The rules concerning all logarithms parallel those we have seen for decimal (base 10) logarithms. Specifically

rLogr Q = Q and Logr rX = X.

Logr (A . B) = Logr A + Logr B (Logr of product equals sum of Logr's)

Logr XA = A Logr X (Logr of power equals power times Logr)

As simple examples, consider the following:

Log2 16 = 4 (since 24 = 16)

Log3 9 = 2 (since 32 = 9)

LogA A = 1 (Why?)

One non-decimal logarithm is so common that it has its own name and notation. (This is the other special case mentioned above.) It is also the other type of logarithm available as a button on your calculator. This is called the natural logarithm and is usually denoted by Ln. The base of natural logarithms is the constant e = 2.71828.... Natural logarithms and exponentials appear in a tremendous number of mathematical models for real world phenomena; you will see them frequently in engineering. Also, you will learn or have already learned in calculus that the constant e is unique because the integral of eX is eX. (This of course implies that the derivative of eX is eX.)

Rephrasing the basic properties for natural logarithms gives

eLn Q = Q and Ln eX = X.

Ln (A . B) = Ln A + Ln B (Ln of product equals sum of Ln's)

Ln XA = A Ln X (Ln of power equals power times Ln)

(Are you sick of seeing those yet?)

Now for some real problems.

In order to solve these problems, you will have to apply your knowledge of logarithms to manipulate the equations into a form that you can either easily estimate or solve on your calculator.

WARNING: The HP48 equation writer function does not always give correct results when dealing with complex logarithmic and/or exponential formulae, so beware. Use the machine as a calculator instead; it is not only more fool-proof, it is faster as well. Some of the following problems you cannot even write with the equation writer, so you will HAVE to modify the statement of the equation.

For now, leave your calculator alone and estimate the answers. You WILL have to think in order to solve some of these, they are not all trivial. Use only the relationships shown earlier in this paper - DO NOT use logarithmic tricks you happen to remember from earlier memorization. DERIVE the relationships you need. Begin learning to think like an engineer.

REMEMBER: don't cheat yourself by looking at the solutions until either you are confident that you know how to do it, or you have honestly tried diligently and are ready to give up in despair.

Question Group C

8. Log X5 + Log X =12; Estimate X

HINT: If you have five aardvarks and add one aardvark, how many aardvarks do you have? Also, remember that 10 Log X = X

9. Log 17X =615; Estimate X

HINT: You might be able to directly estimate this one using your understanding of logarithms, but when you solve it using your calculator, you need to manipulate it into a form where X = {some expression}. Your first task is to get that X moved from its terribly inconvenient location in the exponent.

10. Ln 32.5Y + 1.5 Ln 3 =77; Estimate Y

HINT: Utilize the techniques you employed in problems 8 and 9.

11. Log54876364 = X; Estimate X

HINT: Use the relation r Logr A = A followed by the relation Log XB = B Log X.

12. LogX50 = 3

HINT: See hint to problem 11.

Go to Question Group C answers.

ANSWERS

Question group A (Return to Group A Problems)

1. 2

2. 5

3. My guess is 3.7; you can see how close I came with your calculator if you wish. Using the hint, upper and lower limits are easily seen to be 3 < X < 4.

4. 3

5. 6

6. As in problem 3, upper and lower limits are 0 < X < 1.

My guess is 0.7

Question group B (Return to Group B Problems)

7. A. Log 3592Log (32 . 100) =Log 25 + Log 1001.5 + 2 = 3.5

Since 3592 is a bit larger than 3200, we will revise our estimate upwards a bit. Since 3592 < 4000, we can easily get an upper limit of 3.6 (4000 =22 . 1000).

My estimate is 3.55

B. Log 305 = 5 Log 305 Log 32 = 5 Log 25 = 25 Log 225 . 0.3 = 7.5

Since 30 is a bit less than 32, let's revise the estimate down a bit.

My estimate is 7.4

C. Log 0.0423 = Log (4.23 . 0.01)Log 22 + Log 10-20.6 - 2 = -1.4

Since 4.23 is a bit larger than 4, we need to revise the estimate up a bit.

My estimate is -1.38

Be careful here; the magnitude of the estimate decreases since the value is negative.

Question group C (Return to Group C Problems)

8. Log X5 + Log X =12 5 Log X + Log X = 12 6 Log X =12 Log X =2

Next, raise 10 to the values on each side of the equality to get the final answer.

10Log X = 102 = 100

But 10Log X = X, so X = 100

To check this answer, simply plug 100 into the equation for X and calculate it.

9. Log 17X =615 X Log 17 = 615 X = 615/Log 17615/1.22500

Again, to check this, plug in the value of X and calculate.

10. Ln 32.5Y + 1.5 Ln 3 =77 2.5Y Ln 3 + 1.5 Ln 3 = 77 (2.5Y + 1.5) Ln 3 = 77

Since 3 is just a little larger than e2.71828, Ln 3 is a little larger than 1; I will estimate Ln 31.1

This yields 2.5Y + 1.577/1.1 = 70 2.5Y =70 - 1.5 = 68.5 Y = 68.5/2.5

This gives Y = 27.4

Check this in the same manner as problems 8 and 9.

11. Log54876364 = X 5Log54876364 =5X 4876364 = 5X .

Log 4876364 = X Log 5 X = Log 4876364 / Log 5

Now to estimate the two logarithms.

Log 50.7

Log 4876364Log (4 . 1000000)6.6 This estimate needs to be revised upwards, however, so I will estimate Log 48763646.69

(If the estimate of Log 5 is correct, it has to be less than 6.7; can you explain why?)

Finally, we have X6.69/0.79.6

To check this one, raise 5 to the power 9.6 and see if it equals 4876364. (If you were estimating, it should be fairly close. My calculator gives 5129937, and error of only 5.2%; not too bad for an estimate.

BE SURE YOU UNDERSTAND WHY THAT METHOD OF CHECKING WORKS!

12. LogX50 = 3 XLogX50 = X3 50 = X3 Log 50 = 3 Log X Log X = Log 50 / 3

Estimating Log 50 gives Log X = 1.7 / 30.57

Finally, 10Log X = 100.57 X3.8

That last estimate was derived from the fact that I know that 100.64. See the section on Logs of powers of 2 and use it in reverse.

To check this one raise 3.8 to the power 3. You should get 50. Checking my estimate yields an error of less than 10%, still not too bad.

BE SURE YOU UNDERSTAND WHY THAT METHOD OF CHECKING WORKS!

NOTICE: If you find any errors in this tutorial, PLEASE bring them to my attention as soon as possible. Thanks.

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