logs17

A QUICK AND DIRTY INTRODUCTION TO LOGARITHMS

(Being my 'personal take' on Dan Umbarger's 'Explaining Math Logarithms': www.mathlogarithms.com. Plagiarized, with pleasure, by Ray Andrews. )

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TERMS:

Exponents and powers are exactly the same thing. Log(arithm)s, antilog(arithm)s and bases are just special ways of using exponents. Here's how we talk it:

2^3=8

Read: "2 to the 3d power (2*2*2) equals 8" or: "8 is the 3d power of 2" or: "the log of 8, base 2, is 3" or: "the antilog, base 2, of 3 is 8".

log_2(8)=3

Read: "the logarithm, base 2, of 8, is 3". (In calculus, this equation and the one above are called the 'inverse' of each other.)

8^{(\frac{1}{3})}=2

Read: "8 to the 1/3d power equals 2: or: "the cube root of 8 is 2".

log_8(2)=\frac{1}{3}

Read: "the log, base 8, of 2, is 1/3".

In all of the above, 2 is the base and 3 is the exponent.

Needless to say, all the normal rules that apply to exponents also apply to logs. (You do know how exponents work don't you?):

log(100)=2

log(\frac{1}{10})=-1

log(\sqrt{10})=0.5

log(\frac{1}{\sqrt{10}})=-.05

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BASE:

Logs can be used with any base number and the rules below work so long as you stick with one base. However, except in special circumstances, the base used is almost always either 'Euler's Number' -- represented by the symbol 'e' (an italic lower case 'e'), giving us the so-called 'natural log', written as 'ln(x)' -- or the base is '10', the so-called 'common log', written as 'log(x)'. If no base is shown then it's a common log, otherwise specify the base:

log_5(125)=3

(If you want to belabor things you can also do this:

log_10(1000)=3

or even this:

log_e(10)=2.3025

... but it's no way to make friends, stick with:

ln(x),\;\;\;\;log(x)\;\;\;\;and \;\;\;\;log_b(x)

.)

Scientists and engineers like common logs because they work well intuitively due to the fact that we use base 10 for our number system, thus: 10^3=1000 and: log(1000)=3. Some scientific terms that use common logs are:

- The Richter scale of earthquake intensity, where each magnitude is 10X more powerful than the one before.
- The pH acidity scale, where '3' is 10X more acidic than '4' (that scale works 'backwards' -- lower numbers mean more acidic.)
- The decibel scale of sound intensity.

Natural logs don't seem 'natural' at all really: ln(1000)=6.99077... But there are deep mathematical reasons why using Euler's Number (2.7182... , which is irrational to make matters worse), as the base for logarithms is a good idea. When you take calculus you'll understand how much magic Euler's Number (pronounced 'oiler') holds; it should be called God's Number, it is the key to the universe and calculus is *only* done with natural logs. Also, calculating natural logs is much simpler than common logs. Here's the formula:

ln(x)=\frac{(x-1)^1}{1*x^1}+\frac{(x-1)^2}{2*x^2}+\;...\;\frac{(x-1)^n}{n*x^n}

... where 'n' is however far you want to continue the sum for increasing accuracy. Written as a summation:

ln(x)=\sum_{n=1}^{100}\frac{(x-1)^n}{n*x^n}

... where we are doing 100 iterations of the sum which gives about 6 decimals of accuracy. Common logs are most easily computed by taking natural logs and doing a 'base conversion' as explained below.

As we will see, one example of a 'strange' base involves the decay of a radioactive substance. We compute the rate of decay using the 'half-life' of the substance, and the calculations use the base: 0.5. More about this later.

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A VERY BRIEF HISTORY OF LOGS:

The whole reason logarithms were invented is so that you could multiply two numbers by simply adding their logs and then taking the antilog. Back before calculators this was a huge time saver! Try one like this for yourself by hand (no cheating!):

12345 x 67890 = 838102050

... now with logs:

ln(12345) = 9.42100... (rounded off)
+ ln(67890) = 11.12564... (rounded off)
======================
=20.54664

antilog(20.54664) = 83809...

Not perfect! That's OK, we could get more accuracy if we'd used more digits of the logs instead of rounding them off like I did. Logs were mostly used by engineers who always round things off anyway, the point is to be accurate enough for the job at hand. If you're designing a dam, calculating the concrete you'll need, the result might be rounded off to the nearest 100 cubic meters. (This invites a discussion of 'significant figures' and 'scientific notation', but that's another topic.) Mathematicians, in contrast, almost never round things off: The number 1.414213562 would make most engineers very happy, but for the mathematician only '

\sqrt2

' would be acceptable.

Now, I hear you asking: Why use a calculator to get logs, add the logs and then take the antilog when we could just multiply the numbers? The reason is that before calculators what we had was little booklets that contained log tables. We looked up the logs, added them, looked up the antilog and got the answer much faster than doing the multiplication by hand. Or there was the formerly ubiquitous slide rule which is essentially a rather simple machine for adding logs, but you didn't get much accuracy:

Slide rules: the early calculators - Chalkdust

Slide rules: the early calculators - Chalkdust

... still it was way cool being the fastest and best in your class with one of them, the hot chicks noticed you. This was the way it was done from the early 1600's when log tables were first produced by a guy named Napier until the first commonly available calculators in the mid 1970s -- log tables for accuracy, slide rules for fast results.

Logs can also be used to compute powers and roots. If you want the cube of a number, take the log of the number, multiply it by three and then take the antilog. If you want the cube root of a number, ditto, but divide instead of multiplying. What about roots of negative numbers? Easy, ignore the sign, perform your operation, and then reattach the sign. How about that for solving a problem?

Finally, as mentioned above, we use natural logs (never common logs!) all the time in calculus in a way that can't be substituted with any other operation.

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LOGARITHM RULES:

Definitions:

In everything below 'b' and 'c' will be numbers used as bases. 'm' and 'n' will be exponents or logs (which is the same thing), and 'x' and 'y' will be 'plain numbers'. These definitions can be proven and you are welcome to do it, but in my opinion they stand as true by definition:

This is the basic definition of a logarithm:

If:

b^m=x

then:

m=log_b(x)

Eg: If:

2^4=16

then:

4=log_2(16)

This is a tautology: the log of a number raised to a power is that power. It goes in a circle:

log_b(b^m)=m

Eg:

log_2(2^4)=4

This is even more belabored: a base raised to the power of the log of a number -- that is, the antilog of the number -- is simply that number. Think of it as an alternate definition:

b^{log_b(x)}=x

Eg:

2^{log_2(16)}=16

Put more simply, comparing the above two rules:

ln(e^x)=x

e^{(ln(x)}=x

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RULES FOR WORKING WITH LOGS:

In all the rules below, if no base is shown, any base can be used. For the examples below I use base 2 at first because it makes the answers obvious. Some examples use base 5 and some use the natural log. Most of these rules have no formal proofs here because they are so obviously true -- they follow directly from the definition of what a logarithm is. Also, note that usually it is easier to 'just see' how these rules work than it is to put them into words in a formal definition.

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Product rules:

The original usefulness of logs: to multiply numbers just add their logs and then take the antilog (that is, raise the base to the power of the log). Or: 'The log of a product is the sum of the logs of the original numbers.' (Remembering that the base must not change!):

xy=b^{(log_b(x)+log_b(y))}

Eg:

4*8=2^{(log_2(4)+log_2(8))}=2^{(2+3)}=32

This says exactly the same thing a bit less confusingly:

b^mb^n=b^{(m+n)}

This says exactly the same thing in log-speak:

log(xy)=log(x)+log(y)

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Quotient rules:

If adding logs multiplies numbers, we are not surprised that subtracting logs divides numbers:

\frac{x}{y}=b^{(log_b(x)-log_b(y))}

Eg:

\frac{32}{4}=2^{(log_2(32)-log_2(4))}=2^{(5-2)}=8

This says exactly the same thing a bit less confusingly:

\frac{b^m}{b^n}=b^{(m-n)}

Eg:

\frac{2^5}{2^2}=2^{(5-2)}=8

This says exactly the same thing in log-speak:

log(\frac{x}{y})=log(x)-log(y)

... Make your own example, again, these work with any base.

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Exponent rules:

If adding logs multiplies numbers, we are not surprised that multiplying logs raises numbers to a power:

x^m=b^{(m\;*\;log_b(x))}

Eg:

4^3=2^{(3\;*\;log_2(4))}=2^{(3*2)}=64

Eg:

(4^3)^2=2^{(3\;*\;log_2(4)\;*\;2)}=2^{(3*2*2)}=2^12=4096

As with the other rules, we can say it in log-speak:

log(x^m)=m\;log(x)

Eg:

log(4^3)=3*log(4)

If multiplying logs raises a number to a power, what does dividing logs do? That's right, it takes the root of a number:

x^{(1/m)}=b^{(log_b(x)/m)}

Eg:

64^{(1/3)}=2^{(log_2(64)/3)}=4

In log-speak:

log(b^{(1/m)})=\frac{log(b)}{m}

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Solve for exponent rule:

If:

b^m=x

then:

m=\frac{log(x)}{log(b)}

As mentioned we can use any base we like, so let's work this example using natural logs:

Eg: If:

2^m=8

then:

m=\frac{ln(8)}{ln(2)}=3

But that's not exactly obvious, so let's prove it:

2^m=8

ln(2^m)=ln(8)

m*ln(2)=ln(8)

m=\frac{ln(8)}{ln(2)}=3

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Change from base 'b' to base 'c' rule (where we are converting from a 'normal' base to a 'strange' base):

log_c(x)=\frac{ln(x)}{ln(c)}

Eg:

log_5(125)=\frac{ln(125)}{ln(5)}=3

We can use any base in the fraction so I use the natural log which is as good as any other. This rule is also not obvious so let's prove it: Our starting number is 125 and we want to find its log 'm', base 5:

5^m=125

ln(5^m)=ln(125)

m*ln(5)=ln(125)

m=\frac{ln(125)}{ln(5)}=3

5^3=125

Changing from a 'strange' base to a 'normal' base ('e' here) is easier:

b^m=e^{ln(b^m)}

Eg:

5^3=e^{ln(5^3)}=e^{4.8283}=125

Notice that solving for base and changing base end up as basically the same rule.

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Swap number and base rule:

log_b(x)=\frac{1}{log_x(b)}

Eg:

log_5(125)=\frac{1}{1/3}=3

log_125(5)=\frac{1}{3}

125^{(1/3)}=5

... which is the inverse of:

5^3=125

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MORE ABOUT EULER'S NUMBER:

As mentioned, Euler's number shows up in math and physics almost as often as pi. Here are a few examples:

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The bell curve:

If you're familiar with statistics you know about the bell curve. Its formula is built around 'e'. Graph this if you have the hardware or software:

\frac{e^{(-x^2/2)}}{\sqrt{2\pi}}*10

... (the '10' just steepens the graph to make it look more 'normal')

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The logarithmic spiral:

... is normally computed using 'e' as the base, although other bases work too. In this case I'm not sure what's so special about 'e'. I don't think there is a simple, easy formula for this. In GeoGebra with the grid set to polar coordinates:

r=e^x
s=Curve(r(t) cos(t), r(t) sin(t), t, inside, outside)

... Where 'r' is the radius, 's' in the name of the function that draws the spiral, 't' is the input variable (basically 'x') and 'inside' and 'outside' are it's inner and outer graphing limits.

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The catenary curve:

If you hang a chain (or some other completely un-stiff rope or cable or whatever) between two horizontally separated points, the shape it takes is not a parabola, as most people suppose, it is a 'catenary' and the equation of its graph is:

y=(e^x+e^{-x})/2

A catenary curve can be thought of as 'half way' between a parabola and an ellipse or semicircle if all three curves share the same endpoints and apex.

... the black curve is a semicircle, orange is a catenary, and blue is a parabola.

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Continuous compound interest:

Most of us are familiar with compound interest. You invest $100 at 6% annual interest; in a year you have $106. Except that almost always, interest compounds. That is, at some point you end up collecting interest on the interest too. In the above case, if interest was compounded yearly, after two years you'd have $112.36, not $112, because after the first year you'd start collecting interest on the $6 too. Compounding is usually semiannual or monthly. This means that 6% annual interest is actually computed as 3% every six months or 0.5% per month respectively.

The compound interest formula is:

V=P(1+\frac{R}{c})^{ct}

... Where V is the final value, P is the principal, R is the annual rate of interest, c is the number of compoundings per time period, and t is the time period (almost always a year in financial calculations).

[ That formula is just a specialized case of the exponential increase equation:

F=IR^n

... where 'F' is the final number, 'I' is the initial number, 'R' is the ratio of expansion (if we were talking money, a 5% interest rate gives a ratio of expansion of 1.05), and 'n' is the number of compoundings. Can you convert from one equation to the other? ]

Here's a good practical example of the 'solve for exponent rule': At 5% annual interest, compounded quarterly, how long would it take to double your money?

2=1(1+\frac{0.05}{4})^{4t}

2=1.0125^{4t}

ln(2)=4t*ln(1.0125)

\frac{ln(2)}{4*ln(1.0125)}=t=13.949

Now, it doesn't take long to realize that the more often interest is compounded the faster your account grows because the lag between accumulating interest, and getting interest on that interest, gets shorter. But what if interest was compounded infinitely often? Believe it or not, the formula for that is simpler than the formula for normal compounding and it relies on Euler's Number:

V=Pe^{rt}

Where 'V' is the final value, 'P' is the principal, 'e' is Euler's Number (of course), 'r ' is the rate of interest per time period (0.05, or 5%, annually here), and 't' is the number of time periods. Using continuous compounding on the above investment, solving for 't', we end up with:

2=1(e^{(0.05*t)})

ln(2)=t*ln(e^{0.05})

t=\frac{ln(2)}{ln(e^{0.05})}=13.862943

... We double our money in 13.86 years with continuous compounding instead of 13.94 years with quarterly compounding.

In finance continuous compound interest is rare but in science and engineering we have many situations where 'compound interest' is 'collected' and it is almost always collected continuously. The growth of bacteria is a good example. Or the decay of a radioactive element (in this case the compounding is negative -- the equation is the same, but with negative 'interest').

Now, let's take the normal compounding equation:

V=P(1+\frac{R}{c})^{ct}

... and try to get very close to continuous compounding by making 'c' very large, say 1,000. P=1 and V=2 because all we care about here is doubling our money.

2=1(1+\frac{0.05}{1,000})^{1,000*t}

Skipping the steps (do them yourself if you need the practice), t=13.8632 ... just a tiny bit less than continuous compounding. I'm sure you've figured it out: as 'c' approaches infinity, the answer approaches equality with continuous compounding.

Can we convert the normal compounding equation to the continuous compounding equation? We sure can. First we must make sure to use the same term for the interest rate. Continuous compounding obviously has no 'c' term and to make things equal let's use one year as the time period for both equations. Thus 'r' = 'R'. Let's start with the simplest example: let's have an interest rate of 1 per annum (that is 100%, or doubling every year if we only compounded once per year), and just one time period . What would our actual final value be compounding continuously starting with one dollar? R=1, r=1, t=1, P=1, so:

V=Pe^{rt}

V=1*e^{(1*1)}

V=e^1 = e = 2.718281

... Euler's Number of course. (Interesting is it not, that '100% per annum, compounded annually', is 271% compounded continuously?)

Now the same input but using the standard formula:

V=P(1+\frac{R}{c})^{ct}

2=1(1+\frac{1}{1,000})^{1,000*t}

V=(1+\frac{1}{1,000})^{1,000}=2.7169

... almost Euler's Number! As 'c' approaches infinity (this is what we call a 'limit' in calculus), it IS Euler's Number. In fact this is one of the definitions of Euler's Number (there are several), so we can substitute the expression within the square brackets with 'e':

V=P[(1+\frac{1}{c})^c]^t

... Where 'c' approaches infinity. Where 'r' =1:

V=Pe^{rt}

... QED. Here's a graph showing how quickly the standard formula converges on 'e' as 'c' (the 'x' value in the graph) increases:

But supposing 'r' and 'R' are not 1? Let's do it with a realistic interest rate of 0.05%:

V=(1+\frac{0.05}{1,000})^{1,000}=1.051269

V=e^{0.05}=1.051271

... try a few examples for yourself, it works. As 'c' approaches infinity the two equations become identical. (A formal proof is beyond my abilities.)

BTW, if you double the interest rate, what happens to your return? Double? Using the continuous compounding formula, Try it and see.

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Radioactive decay

... is a good example of negative compound interest.

As mentioned above, for this sort of problem we use the exponential increase equation:

F=IR^n

... where 'F' is the final number, 'I' is the initial number, 'R' is the ratio of expansion and 'n' is the number of compoundings. Let's make an example using the decay of carbon 14, which has a half-life of 5750 years. In other words, every 5750 years the quantity of C14 is cut in half. Thus the ratio of expansion is 0.5. Given an initial quantity of 100 grams, how much C14 do we have after 20,000 years:

F=100*0.5^{(20,000/5750)}=8.9730\; grams

Now, it's rather obvious that the rate of decay is smooth, there is no arbitrary compounding period like semiannually, is there? So it would be more scientific to convert the above equation into one based on 'e' which is more natural for continuous compounding:

ln(0.5)=-0.693147

... so the 'continuous compounding' or 'ratio of expansion' or 'half-life' of C14 in natural log terms is -0.693147. So 'e' raised to that power, and raised again to 20,000/5750 will give us the same result:

F=100*e^{-0.693147*(20,000/5750)}=8.9730\; grams

Is it worth the bother of converting? Yes, because solving certain problems becomes much easier. For example, if an object started with 100 grams of C14, and it now has 90 grams, how old is it?

90=100*e^{(-0.693147*y/5750)}

ln(\frac{90}{100})=ln(e^{-0.693147*y/5750})

... divide both sides by 100 and take the natural log of both sides.

-0.105336=-0.693147*y/5750

... remember that the natural log of 'e' raised to some number is just that number.

-0.105336/-0.693147*5750=y=874 \;years

... try it with the 'half-life' formula above and you'll see it's more difficult because you are working with base 0.5 logarithms which you calculator does not do, so you are forced to convert the base anyway. Base 'e' natural logs work much better.

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THOUGHTS ON PEDAGOGY:

This doc is intended to be my ... reflection? ... of what I learned, or re-learned, from Dan Umbarger's book. Rather than writing a test, I re-write the textbook to show what I've learned. (Some of Dan's book is still too hard for me, but I have the essentials.) Anyway, Dan is devoted to the art, skill and science of pedagogy and his book is built around very well enunciated principles. Read:

www.mathlogarithms.com/note-to-teachers.htm

In particular I note these insights:

> It has always seemed ironic that authors and teachers, so knowledgeable about mathematical sequences, could be so insensitive and clumsy about the sequencing of curriculum …

> As such, I present many, many examples to help the student to see patterns and only then do I present the abstraction which will allow for generalization to all cases. Induction is a powerful teaching tool.

> I believe that the best way to introduce a new idea is to somehow relate it to previous ideas the student has been using for some time. Using this approach, new concepts are an extension of previous ideas

> For example, most high school text books seem to shy away from a meaningful discussion of why scientists and other professionals prefer to work with base e, the natural log, rather than the more intuitive common base, base 10. They do so because the pre-calculus student has not yet been exposed to the ideas that are necessary to justify the use of base e.
-----------------------------------

I'm fascinated by pedagogical principles too, but my thoughts on the subject are as yet more instinctive than enunciated. This doc is a 'response' to the principles exemplified in Dan's book in which I build upon, and slightly modify, the work to fit my own notions in the hope of clarifying for myself what I think works. This is a sort of 'experiment'.

I completely agree with Dan that the 'broad front' advance of math knowledge in schools is inferior to the 'deep dive'. As he says, when discussing logs, we should let students know *why* natural logs are preferred even if the student technically speaking 'doesn't need to know it yet' -- she might still be curious, and various mathematical concepts should not 'dangle' in the mind. Thus, when the subject is logs, let's 'go all the way'. Let's finish what we start rather than smearing out the subject over two or more years.

My doc is informal, personal and conversational. The presentation is as simple as it can be and the style is not rigorous. Rigid definitions are sometimes given, but they are deprecated as not really that important. I don't have any exercises but that's only because I'd want my reader to use Dan's exercises. (And simple laziness -- I've made something only slightly more developed than a cheat-sheet.)

I like digressions that I hope are interesting and might keep the student interested. But one can wander too far afield, this is a matter of skill and judgement.

I like building an equation into a sentence. Not sure if this is good tho.

One thing where I think Dan could have done better is in the ordering of the information. For example, he leaves some definitions of terms to near the end. There's a few places where it's obvious that I've reorganized things. Better?

I'm fascinated by the idea that it might be possible to learn math *much* faster than is normally supposed. This ties in with the 'deep dive' concept -- fire up the forge, get the metal hot and *keep hammering* while it is hot. This is obviously 'dangerous' thinking, but might be fruitful.

Dan emphasizes repetition. It's hard to disagree, yet I harbor the idea that once the student 'gets it' -- *really* gets it -- one need not belabor the matter. But constant, if short, reviews should be emphasized.

I like to 'tease' the student. There's places where I deliberately don't show all the steps in my work, and I invite the student to fill in the gaps. Too much of this would backfire tho.

Something I like that is very rarely done is that I let the student know what's obvious and what's difficult. It is good IMHO to advise a student that the next equation is a doozy, and if he's having trouble with it, that's to be expected -- he's not stupid.

What else?