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Posts tagged ‘logs’

The Log Blog


Tales from the Tutoring Experience

Tales from the Tutoring Experience

It’s that time of year again when Algebra 2 students are all studying logs — not the lumberjack things, but ‘logarithms’ — so I’d like to present a concept that helps students work with logs.

I call it the “Whole-Part-Equality Principle” (as I’ve never seen it named by anyone else … is there a name for it? Anyone know?) But I prefer to call it the “Peter-Paul-Pennies-in-the-Pocket Principle.”

Here’s how it works. There’s Peter, and there’s Paul. We are told that Peter and Paul have no money except pennies, and they transport their pennies only in their right and left pants pockets (if anyone can think of a way to pack this story with even more p’s, please let me know).

Anyhow, we know three additional facts:

1)  The number of pennies that Peter is transporting equals the number of pennies that Paul is transporting.

2) Peter and Paul each have three pennies in their right pants pockets.

3)  Peter and Paul transport their pennies NOWHERE but in their pants pockets.

QUESTION:  What can we conclude about the number of pennies that Peter and Paul have in their left pants pockets?

ANSWER:  It’s obvious, right? While we don’t know how many pennies Peter and Paul could be transporting in their left pants pockets (it could be any number, right?), it is nevertheless clear that they must have the same number of pennies in their left pants pockets.

WHY?  View it like this … If the wholes are the same (the total number of pennies that Peter and Paul each has), and if one of two key parts are the same (the number of pennies that Peter and Paul have in their right pants pockets), then the other parts must also be equal (the number of pennies they have in their left pants pockets).

Why am I bringing this up? To point out an important principle.

This same principle — if the wholes are equal, and if one of their two parts are equal, then the other parts must also be equal — can be used to solve many log and exponent problems.

EXAMPLE 1:  Suppose you have this equation:  log x = log 7.2. What can we conclude? Well, the wholes are equal (meaning the left and right sides of this equation are equal), and the bases of the logs are equal (logs are always base 10 unless another base is given), therefore the remaining parts, the ‘arguments,’ also must be equal. The ‘argument’ is the term after the word ‘log,’ so for this equation the arguments are x and 7.2, and they must be equal … meaning that  x = 7.2.

EXAMPLE 2:   Suppose you have the equation:  log 2^x = log 16. Again, the wholes are equal, and the logs have the same base, so the arguments must be equal. That means that 2^x = 16. Since 2^4 = 16, x = 4, and that’s the answer.

EXAMPLE 3:  Suppose you have the equation:   a^log x = a^log 12.9. Since the wholes (the left and right sides of the equation) are equal, and since the bases are equal as they are both ‘a,’ therefore the only remaining parts, the exponents, must also be equal. So this means that log x = log 12.9. Following the same logic as we used in Examples 1 and 2, this means that x = 12.9.

Any questions? If so, please post as a comment. If not, please use this principle, and enjoy its profound practicality. (OK, I’m done.)

Josh Rappaport is the author of the Algebra Survival Guide and Workbook, which together comprise an award-winning program that makes algebra do-able! Josh also is the author of PreAlgebra Blastoff!, an engaging, hands-on approach to working with integers. All of Josh’s books, published by Singing Turtle Press, are available on Amazon.com

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The Clouds Part, and a Log Rule MAKES SENSE!


Have you ever been befuddled by the rules for logs?

More specifically, have you ever looked at this rule:

log (v w) = log v + log w

and thought: Now why in the world is that true?! What exactly is this saying? I know that I, myself, have had that thought. And for me the desire to understand this rule never went away. Till I got it some time ago.

[By the way, keep in mind that the v and the w in the parentheses are multiplying each other, so that v w actually means: v times w]

And the good news is: I think I can explain this rule in a way so that pretty much everyone who knows basic algebra can grasp it.

O.K., first, I knew that this log rule was related to another rule, the  exponent rule that says:

(a^b) x (a^c) = a^(b + c)

Remember: this is the rule that says if you have two exponential terms  with the same base, and those two terms are multiplying each other, you just keep that base and add the exponents. For example:
(3^2)  x  (3^5) =  3^(2 + 5) = 3^7

But how exactly does this exponent rule relate to the more confusing-looking log rule?

To get ready to see this, one preliminary concept must be clear. The concept is that whenever you see a log term, you’re basically seeing an exponent. Why? Because every log represents an exponent. For example:  log 2 of 8 is the exponent of 3 since 2^3 = 8. 

Put another way, the term log 2 of 8 is asking a question. It’s asking: what exponent would you plunk on the right shoulder of the smaller number, 2, to get the much bigger number 8? The answer is 3, since 2^3 = 8.

Now you try this.

What question is log 3 of 81 asking? Answer: What exponent would we put on 3 to get 81?
What is the answer to this question? Answer:  4, since 3^4 = 81.
So based on all of that, log 3 of 81 = 4.

Now that we’ve got this concept straight, let’s look at the log rule again.

log (v w) = log v + log w

If we substitute in some numbers, this rule will be easier to think about. So let’s substitute 4 for v and 8 for w. After doing that we get:

log (4 x 8) = log 4 + log 8

Next, keep in mind that we can insert a base, and we can actually use any base we wish, as long as we use the same base for all three terms. A handy base would be 2 since 4 and 8 are both powers of 2. So when we use 2 as our base, the equation now reads:

log 2 of (4 x 8) = log 2 of 4 + log 2 of 8

One more thing before we tackle this sucker. Let’s  express the product inside parentheses as 32, which is ok since 4 x 8 equals 32, right? So now the equation reads:

log 2 of (32) = log 2 of 4 + log 2 of 8

Now, after all of that work, let’s finally have some fun. “Having fun,” of course, is relative, but if you’re a math person, “having fun” probably means: let’s  figure out what this crazy equation is saying. So here goes …

Based on what we’ve been saying, the left side of the equation asks the question: what exponent would we put on 2 to get the number 32. So what about that … ? What exponent would we stick on the left shoulder of 2 to get 32? The answer, of course, is 5, since 2^5 = 32. O.K., so far so good: the left side of this equation is clearly equal to 5.

Now how about the right side? While the left side asked one question, the right side asks two questions because it has two log terms. First, the term, log 2 of 4, asks: what exponent do we put on 2 to get the number 4? That, of course, is 2, since 2^2 = 4. And the next term, log 2 of 8, asks: what exponent do we put on 2 to get the number 8? That, of course, is 3, since 2^3 = 8.

So the two log terms on the right side are 2 and 3. And we are supposed to add those terms because the equation says to add them. And what is 2 + 3? It is 5, the same number we just got for the left side of the equation. So that is that. The rule works. We can see it working!

And all it is really saying (for this example) is this:

The exponent you put on 2 to get 32 [which is 5] is the sum of the exponents you put on 2 to get the factors of 32, 4 and 8. Or, stated more succinctly and more generally:  the exponent you put on a base to get a certain number is the sum of the exponents you put on that same base to get the factors of that certain number.

That is all that this formula is saying; nothing more, nothing less. So if you understand what I’ve explained here, you understand this rule more deeply. And that is a cool thing. So pat yourself on the back, and go  enjoy the rest of your day!