We talked a little about Bloglines today which allows you to gather all the RSS feeds that you want and read them all in one place. This blog has RSS feeds for the main content as well as for the comments. Both links are further down on the sidebar.

Also I wanted to mention on the handout from today that when you set up your powers of two scale it should correspond to the powers of ten scale. In other words both power scales are the same except for which values are marked (powers of 10 on the power of ten scale and powers of 2 on the powers of two scale.) The exponent scales will not be lined up except for zero.

Really the only thing that came up in class worth mentioning is that although we have spent the last week on logarithms we spent twice that amount of time on other functions. This means that the test will reflect that by being about two-thirds 7-2, 7-3 and about one-third logarithms. I plan on having each learning target represented.

My only other advice is that you don’t memorize, but you use what you know about the concepts to reason through the problems. Memory can fail you and your problem solving skills are definately more reliable.

We discussed most of the claims for the logarithm problems. We will finish them up tomorrow and talk about the learning targets for Thursday’s mid-quarter exam.

#7.c shed some light on where 7 goes on the power scale. Noting that 7²≈50, means that 2log(7)≈log(50). So then log(50) can be rewritten as log(10⋅5) or log(10) + log(5), both of which we know from previous work. Dividing by 2 will get us log(7) which is 7′s position on the power scale as determined by looking at the exponent scale.

Pretty cool – thanks Ms. Waterski.

Before I forget – today we tried to come up with a way to find exactly where 7 is on the power scale. We had some answers toward the end of class that were not correct but I think we thought they were. For example getting to 28 by doing log(2) + log(14) would not work because we don’t know log(14). Taking log(3) seven times would not get us to 21, it would take us to 3⁷ =2187.

We have to remeber that the relationship on the power scale is multiplicative not additive. So if we want to find 7 on the power scale we need to “create” 7 using only multiplication and division and only the values we know like log(2) through log(6) as well as other ones we can build like log(8/3). So the problem of 7′s exact position on the power scale remains open.

Tomorrow we will present problems from the problem handout that was given out last week. All 13 problems are up for claiming.

We talked about two interpretations of logarithms today. One way to think of them (more of a function interpretation) is in terms of an input/output relationship. Multiplying to get from input to input produces corresponding outputs that differ by a constant. This is a multiply-add relationship. Which, by the way, makes sense since exponential functions display an add-multiply property and exponentials are inverses of logs.

Another way to think about logarithms is in terms of distance. If you picture the two scales (power and exponent) lined up so 0 (on the exponent scale) appears under 1 (on the power scale) and 1 lines up under 10 and so on, then the distance from 0 to whatever power number you want is equal to the logarithm of that power number.

We ended our discussion with the idea that multiplying power numbers is the same as adding distances on the exponent scale. This means (as usual!) that multiplying powers results in adding exponents (and those exponents are logarithms).