September 24, 2006

Scribe Post: Day 11

Good day to you all, I'm Christian, your scribe. Sorry I wasn't able
to post this right away. There were very important matters I had to
attend to AND I forgot how to put posts up. Thanks MARK for
reminding me how. Thanks to LINDSAY and CHARLOTTE too for teaching me how to put exponents and subscripts here. I'll go on with our lesson shortly.


Over the week we learned about functions, inverse functions. It is
easy most of the time to solve for variables because we only need
to use simple operations. For example, when solving for x in the
(x + 5) / 2 = 6, we:

Step 1: Multiply both sides by 2
x + 5 = 12

Step 2: Subtract 5 from both sides
x = 7

In this case we only used multiplication and subtraction to solve for x. What happens if the variable we're looking for is an exponent? We can use our basic operations as well, but first we must understand what logarithms are, and when we should use them. Logarithms are exponents. When we are solving for an exponent in an equation, we are essentially solving the logarithm for that exponent. Look at the general equation for logarithms:

Logba = c

For this one, b represents the base, a represents the power, and c the exponent. Notice that there is an equal sign between the log expression and c. What does that mean? LOGS ARE EXPONENTS! If logarithms are exponents, to work with a log, we must understand this: In theory, since logs are exponents, the rules of logs and exponents must be the same right? Yes. The rules of logarithms describe what's happening to the exponents. Hence we must understand the rules of exponents.

We learned these rules in grade 12 pre-calculus. This will just be a quick review. As a note again, and to make things clear here, I followed up the "cases" with the general equation in logarithms. As I mentioned, the rules of logarithms describe what happens to the exponents. We simply apply the exponent rules, and from there we can already manipulate our logs.
Case 1: When we multiply terms with the same base, we add the exponents.

(bx)(by) = bx+y

Let's apply this to some value:

This rule states that:
(34)(35) = 34+5
= 39

In logs, this means that:
x + logy = log(xy)

Case 2: When we divide terms with the same base, we subtract the exponents.

(bx) / (by) = bx-y

Again, applying to specific values, this rule states that:

(35) / (34) = 35-4
= 31

In logs, this means that:

log(x/y) = logx - logy

Case 3: When we have a base raised to an exponent and the whole term is raised to another exponent, we multiply the exponents.

(bx)y = b(x)(y)
= bxy

The rule states that:

(3^4)5 = 3(4)(5)
= 320

In logs, this means that:

logb(x)y = (logbx)(logby)

= ylogb(x)

Case 4: The Change of Base Law. This law states that if we have a log written in any base, we can switch the bases. We can easily solve logs of base "10" and base "e", because these are programmed in our calculator.

Take out your calculator. Turn it on. Clear the screen. If we press the log button,
log( will come out right? What we do next is we enter the power, close the bracket, enter, and we get our value. What do you notice in terms of the structure of a log? We did NOT enter the base! What happens then if we are given bases of 2, 900, or 3.2562435234? We apply the Change of Base Law.

This law states that:

logba = log(a) / log(b)

For example:

log34 = log(4) / log (3)

YAY! Then we had bus ridership, and that was our class. Sorry again for posting late. Please visit this.

The next scribe is Anh.

Feel free to add and comment to this. Be nice, though, I have a very fragile ego.


christian said...

I FORGOT TO CHOOSE WHO THE NEXT SCRIBE SHOULD BE!!! Again, apologies. I can't find the edit button so I don't know how to change my post now. I'd just work with the assumption that when people see this post, they'd look at the comments too. So I guess I'll just mention here who it is. The next scribe is ANH.

lindsay said...

good job on your post christian! i needed a review of those log laws. i liked how you used colour in a meaningful way. it wasn't over the top and easy to understand! it was very organized which made it easy to follow. yay!! 1000 points for you christian!

be happy.

linger said...

nice post. yes it was up late, but it's nicely done =) thanks for explaining it well. your explanation was clear, easy to understand and to follow. there's nothing confusing about logs on your post... one thing i have to add; in case 3 where you put(b^x)y = b ^(x)(y) , shouldn't it be b^(x) * b^(y)= b^(x)(y) ? so you're just missing a b before the first y, but other than that good job, keep it up ;)

linger said...

haha ok i just made a mistake. forget about the part of my comment with the b and the x's and y's. i just realized that the y is an exponent, so you don't need another b. oh myyy... ok well that's all -_- "

jann said...

Super job on the post Christian! Your post was well explained. It would really help me in understanding logs. Good job!