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Archive for the ‘Formulas’ Category

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!

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Challenge Problem – Polygon Diagonal Formula


Here’s a challenge problem for anyone who’d like to try it.

On Monday I will post the answer and the names of the first five people who got this right. So good luck, everyone.

A regular polygon is a polygon all of whose sides are congruent and all of whose  angles are congruent. For any polygon, a “diagonal” is defined as a line segment that runs from one vertex of the polygon to another, and which runs through the interior of the polygon.

Find a formula that tells how to determine the number of diagonals there are in any regular convex polygon with n sides.

Once you have the formula, use it to figure out the number of diagonals in a regular convex polygon with 1,000 sides (don’t try this by hand! — that’s why algebra was invented).

Good luck!